Cách làm toán dạng tìm x lớp 6 năm 2024

Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:

\[\begin{array}{l}1]x + a = b \Rightarrow x = b - a\\2]x - a = b \Rightarrow x = b + a\\3]a - x = b \Rightarrow x = a - b\\4]a.x = b \Rightarrow x = \dfrac{b}{a}\\5]a:x = b \Rightarrow x = \dfrac{a}{b}\\6]x:a = b \Rightarrow x = a.b\\7]\dfrac{a}{b} = \dfrac{x}{c} \Rightarrow x = \dfrac{{a.c}}{b}\\8]{x^2} = {a^2} \Rightarrow \left[ {\begin{array}{*{20}{c}}{x = a}\\{x = - a}\end{array}} \right.\\9]{x^3} = {a^3} \Rightarrow x = a\end{array}\]

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Bài tập

Bài 1:

Tìm x, biết:

\[\begin{array}{l}a]2x + 3 = 1\dfrac{2}{3}\\b]0,15 - 3x = {[ - 10]^0}\\c] - x:\dfrac{2}{5} = 0,8\\d]\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\e]\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\f]\left[ {x + 1} \right].\left[ { - 2x - 3} \right] = 0\end{array}\]

Bài 2:

Tìm \[x\], biết:

1. \[\dfrac{1}{3}x + \dfrac{2}{5}\left[ {x - 1} \right] = 0\]

1. \[3 \cdot {\left[ {3x - \dfrac{1}{2}} \right]^3} + \dfrac{1}{9} = 0\]

1. \[3 \cdot \left[ {1 - \dfrac{1}{2}} \right] - 5 \cdot \left[ {x + \dfrac{3}{5}} \right] = {\rm{ \;}} - x + \dfrac{1}{5}\]

1. \[\dfrac{{3 - x}}{{5 - x}} = {\left[ {\dfrac{{ - 3}}{5}} \right]^2}\]

1. \[x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\]

1. \[\left[ {\dfrac{7}{5}\; + \;x} \right]:\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\]

1. \[ - 4:\left[ {x + \dfrac{{ - 2}}{3}} \right] = \dfrac{3}{4}\]

1. \[\left[ {\dfrac{{ - 1}}{5} + 2} \right]:\left[ {x - \dfrac{7}{{10}}} \right] = \dfrac{{ - 1}}{4}\]

Bài 3:

Tìm tập hợp các số nguyên x để: \[\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}\]

Lời giải chi tiết:

Bài 1:

Tìm x, biết:

\[\begin{array}{l}a]2x + 3 = 1\dfrac{2}{3}\\b]0,15 - 3x = {[ - 10]^0}\\c] - x:\dfrac{2}{5} = 0,8\\d]\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\e]\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\f]\left[ {x + 1} \right].\left[ { - 2x - 3} \right] = 0\end{array}\]

Phương pháp

Áp dụng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x.

Lời giải

\[\begin{array}{l}a]2x + 3 = 1\dfrac{2}{3}\\2x + 3 = \dfrac{5}{3}\\2x = \dfrac{5}{3} - 3\\2x = \dfrac{5}{3} - \dfrac{9}{3}\\2x = \dfrac{{ - 4}}{3}\\x = \dfrac{{ - 4}}{3}:2\\x = \dfrac{{ - 4}}{3}.\dfrac{1}{2}\\x = \dfrac{{ - 2}}{3}\end{array}\]

Vậy \[x = \dfrac{{ - 2}}{3}\]

\[\begin{array}{l}b]0,15 - 3x = {[ - 10]^0}\\0,15 - 3x = 1\\3x = 0,15 - 1\\3x = 0,85\\3x = \dfrac{{17}}{{20}}\\x = \dfrac{{17}}{{20}}:3\\x = \dfrac{{17}}{{20}}.\dfrac{1}{3}\\x = \dfrac{{17}}{{60}}\end{array}\]

Vậy \[x = \dfrac{{17}}{{60}}\]

\[\begin{array}{l}c] - x:\dfrac{2}{5} = 0,8\\ - x:0.4 = 0,8\\ - x = 0,8.0,4\\ - x = 0,32\\x = - 0,32\end{array}\]

Vậy x = -0,32

\[\begin{array}{l}d]\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\5.[3x + 2] = 3.[ - 4]\\15x + 10 = - 12\\15x = - 12 - 10\\15x = - 22\\x = \dfrac{{ - 22}}{{15}}\end{array}\]

Vậy \[x = \dfrac{{ - 22}}{{15}}\]

\[\begin{array}{l}e]\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\\left[ {3x + 2} \right].\left[ {3x + 2} \right] = [ - 8].[ - 2]\\{\left[ {3x + 2} \right]^2} = 16\\{\left[ {3x + 2} \right]^2} = {4^2}\\\left[ {\begin{array}{*{20}{c}}{3x + 2 = 4}\\{3x + 2 = - 4}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{3x = 2}\\{3x = - 6}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{x = \dfrac{2}{3}}\\{x = - 2}\end{array}} \right.\end{array}\]

Vậy \[x \in \left\{ {\dfrac{2}{3}; - 2} \right\}\]

\[\begin{array}{l}f]\left[ {x + 1} \right].\left[ { - 2x - 3} \right] = 0\\\left[ {\begin{array}{*{20}{c}}{x + 1 = 0}\\{ - 2x - 3 = 0}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{x = - 1}\\{x = \dfrac{{ - 3}}{2}}\end{array}} \right.\end{array}\]

Vậy \[x \in \left\{ { - 1;\dfrac{{ - 3}}{2}} \right\}\]

Bài 2:

Tìm \[x\], biết:

1. \[\dfrac{1}{3}x + \dfrac{2}{5}\left[ {x - 1} \right] = 0\]

1. \[3 \cdot {\left[ {3x - \dfrac{1}{2}} \right]^3} + \dfrac{1}{9} = 0\]

1. \[3 \cdot \left[ {1 - \dfrac{1}{2}} \right] - 5 \cdot \left[ {x + \dfrac{3}{5}} \right] = {\rm{ \;}} - x + \dfrac{1}{5}\]

1. \[\dfrac{{3 - x}}{{5 - x}} = {\left[ {\dfrac{{ - 3}}{5}} \right]^2}\]

1. \[x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\]

1. \[\left[ {\dfrac{7}{5}\; + \;x} \right]:\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\]

1. \[ - 4:\left[ {x + \dfrac{{ - 2}}{3}} \right] = \dfrac{3}{4}\]

1. \[\left[ {\dfrac{{ - 1}}{5} + 2} \right]:\left[ {x - \dfrac{7}{{10}}} \right] = \dfrac{{ - 1}}{4}\]

Phương pháp

Áp dụng các qui tắc cộng, trừ, nhân, chia phân số, qui tắc tính giá trị của biểu thức.

Lời giải

1. \[\dfrac{1}{3}x + \dfrac{2}{5}\left[ {x - 1} \right] = 0\]

\[\begin{array}{l}\dfrac{1}{3}x + \dfrac{2}{5}x - \dfrac{2}{5} = 0\\\left[ {\dfrac{1}{3} + \dfrac{2}{5}} \right]x = \dfrac{2}{5}\\\dfrac{{11}}{{15}}x = \dfrac{2}{5}\end{array}\]

\[x = \dfrac{2}{5}:\dfrac{{11}}{{15}}\]

\[\begin{array}{l}x = \dfrac{2}{5} \cdot \dfrac{{15}}{{11}}\\x = \dfrac{6}{{11}}\end{array}\]

Vậy \[x = \dfrac{6}{{11}} \cdot \]

1. \[3.{\left[ {3x - \dfrac{1}{2}} \right]^3} + \dfrac{1}{9} = 0\]

\[\begin{array}{l}3.{\left[ {3x - \dfrac{1}{2}} \right]^3} = - \dfrac{1}{9}\\{\left[ {3x - \dfrac{1}{2}} \right]^3} = - \dfrac{1}{9}:3\\{\left[ {3x - \dfrac{1}{2}} \right]^3} = - \dfrac{1}{{27}} = \left[ {\dfrac{{ - 1}}{3}} \right]\end{array}\]

\[ \Rightarrow 3x - \dfrac{1}{2} = {\dfrac{{ - 1}}{3}^3}\]

\[\begin{array}{l}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{{ - 1}}{3} + \dfrac{1}{2}{\kern 1pt} {\kern 1pt} \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{{ - 2}}{6} + \dfrac{3}{6}\\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{1}{6}{\kern 1pt} {\kern 1pt} \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = \dfrac{1}{{18}}\end{array}\]

Vậy \[x = \dfrac{1}{{18}} \cdot \]

1. \[3.\left[ {1 - \dfrac{1}{2}} \right] - 5\left[ {x + \dfrac{3}{5}} \right] = {\rm{ \;}} - x + \dfrac{1}{5}\]

\[\begin{array}{*{20}{l}}{3 - \dfrac{3}{2} - \left[ {5x + 5.\dfrac{3}{5}} \right] = {\rm{ \;}} - x + \dfrac{1}{5}}\\{\dfrac{3}{2} - 5x - 3 = {\rm{ \;}} - x + \dfrac{1}{5}}\\{ - 5x + x = \dfrac{1}{5} - \dfrac{3}{2} + 3}\end{array}\]

\[\begin{array}{*{20}{l}}{ - 4x = \dfrac{{ - 13}}{{10}} + 3}\\{ - 4x = \dfrac{{17}}{{10}}}\\{x = \dfrac{{17}}{{10}}:\left[ { - 4} \right]}\\{x = {\rm{ \;}} - \dfrac{{17}}{{40}}}\end{array}\]

Vậy \[x = {\rm{ \;}} - \dfrac{{17}}{{40}} \cdot \]

1. \[\dfrac{{3 - x}}{{5 - x}} = {\left[ {\dfrac{{ - 3}}{5}} \right]^2}\]

Điều kiện: \[5 - x \ne 0 \Leftrightarrow x \ne 5.\]

\[\begin{array}{*{20}{l}}{ \Rightarrow \dfrac{{3 - x}}{{5 - x}} = \dfrac{9}{{25}}}\\{ \Rightarrow \left[ {3 - x} \right].25 = 9.\left[ {5 - x} \right]}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 75 - 25x = 45 - 9x{\kern 1pt} }\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 25x + 9x = 45 - 75}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 16x = {\rm{ \;}} - 30}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = \dfrac{{ - 30}}{{ - 16}} = \dfrac{{15}}{8}}\end{array}\]

Vậy \[x = \dfrac{{15}}{8} \cdot \]

1. \[x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\]

\[\begin{array}{*{20}{l}}{x:\dfrac{5}{8} = \dfrac{1}{7}}\\{x\;\;\;\;\; = \dfrac{1}{7} \cdot \dfrac{5}{8}}\\{x\;\;\;\;\; = \dfrac{5}{{56}}.}\end{array}\]

Vậy \[x = \dfrac{5}{{56}}\]

1. \[\left[ {\dfrac{7}{5}\; + \;x} \right]:\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\]

\[\begin{array}{*{20}{l}}{\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}}\\{ \Rightarrow \dfrac{{ - 1}}{{24}} \le \dfrac{x}{{24}} \le \dfrac{5}{{24}}}\\{ \Rightarrow {\rm{ \;}} - 1 \le x \le 5}\end{array}\]