giúp mình vớiiii

`#040911`

a,

\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

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

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)

d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)

a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)

\(\Leftrightarrow-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5-\frac{2}{3}\)

\(\Leftrightarrow-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=\frac{13}{3}\)

\(\Leftrightarrow-\frac{1}{3}\left(x-\frac{3}{2}\right).6-\frac{1}{2}\left(2x-1\right).6=\frac{13}{3}.6\)

\(\Leftrightarrow-2\left(x-\frac{3}{2}\right)-2\left(2x+1\right)=26\)

\(\Leftrightarrow-8x=26\)

\(\Leftrightarrow x=\frac{26}{-8}=\frac{13}{-4}\)

\(\Rightarrow x=-\frac{13}{4}\)

b) \(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

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

\(\Leftrightarrow\frac{1}{3}.x+\frac{2}{5}-x-1=0\)

\(\Leftrightarrow\frac{x}{3}+\frac{2}{5}-x-1=0\)

\(\Leftrightarrow-\frac{2x}{3}=\frac{3}{5}\)

\(\Leftrightarrow x=\frac{3}{5}:-\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{9}{10}\)

\(\Rightarrow x=-\frac{9}{10}\)

2.

a. \(A=\left(a+b-c\right)-\left(2a+b-2c\right)\)

\(=a+b-c-2a-b+2c\)

\(=-a+c\)

Thay a=-1 ; c=1 vào A ta có:

\(A=-\left(-1\right)+1=1+1=2\)

Vậy A = 2 với a=-1 ; c = 1

b. \(B=a-\left[\left(a-3\right)+\left(a+3\right)-\left(a-2\right)\right]\)

\(=a-\left(a-3+a+3-a+2\right)\)

\(=a-a+3-a-3+a-2\)

\(=\left(a-a-a+a\right)+\left(3-3-2\right)\)

\(=-2\)

Vậy B = -2

a: =>(3/2-2x):2/3=1/6

=>3/2-2x=1/6x2/3=2/18=1/9

=>2x=25/18

hay x=25/36

b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)

=>x-1/4=1/2

=>x=3/4

c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)

=>23/12x=2/3

=>x=8/23