a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

giúp mik ☺

Giá trị của biểu thức  2 x - 3 35 + x x - 2 7  không lớn hơn giá trị của biểu thức  x 2 7 - 2 x - 3 5  nghĩa là  2 x - 3 35 + x x - 2 7   ≤   x 2 7 - 2 x - 3 5

Ta có:

⇔ 2x – 3 + 5 x 2  – 10x  ≤  5 x 2  – 14x + 21

⇔ 2x + 5 x 2 – 10x – 5 x 2  + 14x  ≤  21 + 3

⇔ 6x  ≤  24

⇔ x  ≤  4

Vậy với x  ≤  4 thì giá trị của biểu thức  2 x - 3 35 + x x - 2 7  không lớn hơn giá trị của biểu thức  x 2 7 - 2 x - 3 5

\(a,B=4x^2+20x+25-9+x^2+14=5x^2+20x+30\\ b,B=5\left(x^2+4x+4\right)+10\\ B=5\left(x+2\right)^2+10\ge10>0,\forall x\)

Do đó B luôn dương với mọi x

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

\(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}=\)

\(=\frac{x\left(x^2+2x\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\)

\(=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x^2-1+4\left(x-1\right)\right)}{2x\left(x+5\right)}=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}\)

a/ Để biểu thức xác đinh => 2x(x+5) khác 0 => x khác 0 và x khác -5

b/ Gọi biểu thức là A. Rút gọn A ta được: 

\(A=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\left(x\ne0;x\ne-5\right)\)

A=1 => x-1=2 => x=3

c/ A=-1/2 <=> x-1=-1 => x=0

d/ A=-3 <=> x-1=-6  => x=-5

a)  ĐK : \(x\ne1\); \(x\ne-1\)

b) Ta có biểu thức:

\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\left(\frac{4x^2-4}{5}\right)\)

\(=\left(\frac{x+1}{2.\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2.\left(x+1\right)}\right).\left(\frac{4.\left(x^2-1\right)}{5}\right)\)

\(=\frac{\left(x+1\right)^2+3.2-\left(x+3\right)\left(x-1\right)}{2.\left(x-1\right)\left(x+1\right)}.\frac{4.\left(x+1\right)\left(x-1\right)}{5}\)

\(=\frac{x^2+2x+2+6-x^2-2x+3}{2.\left(x-1\right)\left(x+1\right)}.\frac{4.\left(x+1\right)\left(x-1\right)}{5}=\frac{40.\left(x+1\right)\left(x-1\right)}{10.\left(x+1\right)\left(x-1\right)}=4\)

Vậy giá trị của biểu thức B không phụ thuộc vào biến x khi  \(x\ne1;x\ne-1\)

\(\left|2x-1\right|=\dfrac{3}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-1=\dfrac{3}{2}\\2x-1=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Thay \(x=\dfrac{5}{4}\) vào D ta có:

\(D=4x+3=4.\dfrac{5}{4}+3=5+3=8\)

Thay \(x=-\dfrac{1}{4}\) vào D ta có:

\(D=4.\dfrac{-1}{4}+3=-1+3=2\)

Để \(D=\dfrac{3}{2}\)

\(\Leftrightarrow4x+3=\dfrac{3}{2}\\ \Leftrightarrow4x=-\dfrac{3}{2}\\ \Leftrightarrow x=-\dfrac{3}{8}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{-1}{4}\end{matrix}\right.\)

\(x=\dfrac{5}{4}\text{ thì }D=4.\dfrac{5}{4}+3=5+3=8\)

\(x=\dfrac{-1}{4}\text{ thì }D=4.\left(\dfrac{-1}{4}\right)+3=\left(-1\right)+3=2\)

\(D=\dfrac{-5}{2}\Leftrightarrow4x+3=\dfrac{-5}{2}\)

               \(\Leftrightarrow4x=\dfrac{-11}{2}\)

               \(\Leftrightarrow x=\dfrac{-11}{8}\)