cho mình xin đề bài

\(\dfrac{3}{2}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=-\dfrac{43}{35}\)

\(\Leftrightarrow x=-\dfrac{86}{105}\)

Vậy \(x=-\dfrac{86}{105}\)

\(-\dfrac{11}{12}x+0,25=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x+\dfrac{1}{4}=\dfrac{5}{6}\)

\(\Leftrightarrow-\dfrac{11}{12}x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{7}{11}\)

Vậy \(x=-\dfrac{7}{11}\)

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

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

Vậy x = {3; 1}\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\)

       (x - 2)² = 1

<=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x = 3; 1

(2x - 1)³ = -8

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

\(|\) 2,5 - x\(|\) = 1,3

1,6- \(|\)x - 0,2 \(|\)= 0

13^x = 169

\(\dfrac{-2}{x}\)=\(\dfrac{-x}{8\dfrac{ }{25}}\)

\(\dfrac{x}{-15}\)=\(\dfrac{-60}{x}\)

\(|\)x\(|\)+0,573=2

2/3 : 0,5 = 2/3 x 2 = 4/3

2/3 : 0,25 = 2/3 x 4 = 8/3

3/8 : 0,5 = 3/8 x 2 = 3/4

9/16 : 0,25 = 9/16 x 4 = 9/4

2/3 : 0,5 = 4/3 

2/3 : 0,25 = 8/3

3/8 : 0,5 = 3/4 

9/16 : 0,25 = 9/4

\(0,25.2\frac{1}{3}.30.0,5.\frac{8}{45}\)

\(=\frac{1}{4}.\frac{7}{3}.30.\frac{1}{2}.\frac{8}{45}\)

\(=\frac{14}{9}\)

\(0,25.2\frac{1}{3}.30.0,5.\frac{8}{45}\)

\(=0,25.\frac{7}{3}.30.0,5.\frac{8}{45}\)

\(=\frac{7}{12}.30.\frac{4}{45}\)

\(\frac{35}{2}.\frac{4}{45}\)

\(=\frac{14}{9}\)

Ta có :3|1/4 - x²| - 0,25 = |x² - 1/4| + 3/8

=> 3|1/4 - x²| - |x² - 1/4| = 3/8 + 0,25

<=> 3|1/4 - x²| - |1/4 - x²| = 5/8

=> 2|1/4 - x²| = 5/8

=> |1/4 - x²| = 5/16

=> \(\orbr{\begin{cases}\frac{1}{4}-x^2=\frac{5}{16}\\\frac{1}{4}-x^2=-\frac{5}{16}\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=\frac{-1}{16}\left(\text{loại}\right)\\x^2=\frac{9}{16}\end{cases}}\Rightarrow x^2=\frac{9}{16}\Rightarrow x^2=\left(\frac{3}{4}\right)^2\Rightarrow x=\pm\frac{3}{4}\)

`25%+3/4+1/2:0,5-1/4:0,15+1/8:0,125`

`=1/4+3/4+1/2xx2-1/4xx4+1/8xx8`

`(1/4+3/4)+(1/2xx2)-(1/4xx4)+(1/8xx8)`

`=1+1-1+1`

`=2`

c) \(\dfrac{x-1}{5}+x=\dfrac{x+1}{7}\)

\(\Leftrightarrow\dfrac{7x-7}{35}+\dfrac{35x}{35}=\dfrac{5x+5}{35}\)

\(\Rightarrow7x-7+35x=5x+5\)

\(\Leftrightarrow7x+35x-5x=5+7\)

\(\Leftrightarrow37x=12\)

\(\Leftrightarrow x=\dfrac{12}{37}\)

Vậy pt có nghiệm duy nhất \(x=\dfrac{12}{37}\)

d) \(2\left(x-2,5\right)=0,25+\dfrac{4x-3}{8}\)

\(\Leftrightarrow\dfrac{16\left(x-2,5\right)}{8}=\dfrac{2}{8}+\dfrac{4x-3}{8}\)

\(\Rightarrow16x-40=2+4x-3\)

\(\Leftrightarrow16x-4x=2-3+40\)

\(\Leftrightarrow12x=39\)

\(\Leftrightarrow x=3,25\)

Vậy pt có nghiệm duy nhất \(x=3,25\)

\(A=\frac{1}{2}:0,25-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)

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

\(A=1+1+1\)

\(A=1.3\)

\(A=3\)

Vậy \(A=3\)