\(\left(1-2x\right)^4=\dfrac{1}{128}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{\sqrt[4]{2}}{4}\\2x-1=\dfrac{-\sqrt[4]{2}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt[4]{2}+1}{8}\\x=\dfrac{-\sqrt[4]{2}+1}{8}\end{matrix}\right.\)

=>100x=128

hay x=1,28

\(\left(4x-5\right)\left(2x+30\right)-4\left(x+2\right)\left(2x-1\right)+\left(10x+7\right)\)

\(=8x^2+110x-150-8x^2-12x+8+10x+7\)

\(=108x-135\)

$(4x-5)(2x+30)-4(x+2)(2x-1)+(10x+7)\\=4x(2x+30)-5(2x+30)-4[x(2x-1)+2(2x-1)]+10x+7\\=8x^2+120x-10x-150-4[2x^2-x+4x-2]+10x+7\\=8x^2+120x-143-4[2x^2+3x-2]\\=8x^2+120x-143-8x^2-12x+8\\=108x-135$

DỂ QUÁ!!!!!!!!!!!!!!!!

TUI HK BIẾT LÀM

\(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\)   \(3\)

<=> \(-4\left(x^2-2x+1\right)+4x^2-1=-3\)

<=> \(-4x^2+8x-4+4x^2-1=-3\)

<=> \(8x-5=-3\)

<=>  \(8x=2\)

<=> \(x=\frac{1}{4}\)

        \(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\)

\(\Rightarrow-4\left(x^2-2x+1\right)+4x^2-1=-3\)

\(\Rightarrow-4x^2+8x-4+4x^2-1=-3\)

\(\Rightarrow8x-5=-3\)

\(\Rightarrow8x=2\Rightarrow x=0,25\)

Vậy \(x=0,25\)

Chúc bạn học tốt.

`x - 3/3 = 4 - 1 - 2x/5`

`->` `x = (-5)`

\(\dfrac{x-3}{3}=4-\dfrac{1-2x}{5}\)

=>5(x-3)=60-3(1-2x)

=>5x-15=60-3+6x

=>5x-15=6x+57

=>6x+57=5x-15

hay x=-72(nhận)

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

\(x=\dfrac{-1}{3}\)

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3

b, 3/4 + 2 . ( 2x - 3 ) = 0 

             2 . ( 2x - 3 ) = 0 - 3/4

             2 . ( 2x - 3 ) = -3/4

                     2x - 3 = -3/4 : 2

                     2x - 3 = -3/8

                          2x = -3/8 + 3

                          2x = 21/8

                            x = 21/8 : 2

                            x = 21/16

Đề bài yêu cầu gì?