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)

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

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

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

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

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

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

\(\Leftrightarrow10x+10\)

a ) A = |2x - 1| - (x - 5)

Ta có : \(\left|2x-1\right|=\hept{\begin{cases}2x-1\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\\-\left(2x-1\right)\Leftrightarrow2x-1< 0\Rightarrow x< \frac{1}{2}\end{cases}}\)

TH1 : 2x - 1 ≥ 0 thì A = 2x - 1 - (x - 5) = 2x - 1 - x + 5 = x + 4

TH2 : 2x - 1 < 0 thì A = - 2x + 1 - x + 5 = - 3x + 6

b ) Để A = 4 <=> x + 4 = 4 hoặc - 3x + 6 = 4 

TH1 : x + 4 = 4 => x = 0

TH2 : - 3x + 6 = 4 => x = 2/3

Vậy x = { 0;2/3 } thì A = 4

a,    A=|2x-1|-(x-5)

      A=|2x-1|-x+5

     A=2x-1-x+5

     A=2x-x+4

     A=x+4

A=2

x+4=2(vì A=x+4)

x    =2-4

x   =-2

a, Để A có nghĩa 

\(\Leftrightarrow x^2-1\ne0\)

\(\Rightarrow\orbr{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

\(b,A=\frac{\left(x-1\right)\left(x-3\right)}{x^2-1}\)

\(A=\frac{\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x-1\right)}\)

\(A=\frac{\left(x-3\right)}{\left(x+1\right)}\)

cảm ơn <3

c, Thay x = 5 vaò A 

\(\Rightarrow A=\frac{5-3}{5+1}\)

\(A=\frac{2}{6}=\frac{1}{3}\)

KL : Giá trị của A = 1/3 tại x = 5

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}\\ A=\dfrac{2x^2+32}{2\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+16}{x^2-4}\\ c,A=-3\Leftrightarrow-3x^2+12=x^2+16\\ \Leftrightarrow4x^2=-4\Leftrightarrow x\in\varnothing\)

Xem lại biểu thức P.

Mình phải đi ăn nên chiều mình làm nốt câu d nhé

a) Điều kiện để P được xác định là: \(x\ne1;x\ne-1\)

b) \(P=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}x-\dfrac{x^2-1}{x^2+2x+1}\)

\(P=\left(\dfrac{\left(x+1\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\right):\dfrac{2x}{5x-5}x-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)

\(P=0:\dfrac{2x}{5x-5}x-\dfrac{x-1}{x+1}\)

\(P=-\dfrac{x-1}{x+1}\)

c) Theo đề ta có:

\(P=2\)

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

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

\(\Leftrightarrow-x-2x=2-1\)

\(\Leftrightarrow-3x=1\)

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

d) \(P=-\dfrac{x-1}{x+1}\) nguyên khi:

\(\Leftrightarrow x-1⋮-\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)-2⋮-\left(x+1\right)\)

\(\Leftrightarrow-2⋮-\left(x+1\right)\)

\(\Leftrightarrow2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)\)

Vậy \(P\) nguyên khi \(x\in\left\{-2;0;-3;1\right\}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

giúp mình với 

Ta có : \(f\left(x\right)=\left|x-1\right|-\left(2x-5\right)\)

Xét 2 TH:

+) Nếu \(\left|x-1\right|=x-1\)

=> \(f\left(x\right)=x-1-2x+5\)

=> \(f\left(x\right)=4-x\)

+) Nếu \(\left|x-1\right|=1-x\)

=> \(f\left(x\right)=1-x-2x+5\)

=> \(f\left(x\right)=6-3x\)

Vậy...

b) \(f\left(5\right)=\left|5-1\right|-\left(2.5-5\right)\)

=> \(f\left(5\right)=4-2=2\)

Vậy...

c) \(f\left(x\right)=0\)

=> \(\left|x-1\right|-\left(2x-5\right)=0\)

=> \(\left|x-1\right|=2x-5\)

Vì \(\left|x-1\right|\ge0\forall x\)

=> \(2x-5\ge0\)

=> \(x\ge\frac{5}{2}\)

=> \(x-1\ge\frac{5}{2}-1=\frac{3}{2}>0\)

=> \(\left|x-1\right|=x-1\)

=> \(x-1-2x+5=0\)

=> \(4-x=0\)

=> \(x=4\)

thank you

a) A = 3x - 4 - ( 2x - 1) = 3x - 4 - 2x + 1 = x - 3

b) Ta có:  x - 3 = 10 => x = 13

ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)

với ĐKXĐ ta có

=\(\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{7\left(x-1\right)}\)

=\(\frac{4x}{\left(x+1\right)\left(x-1\right)}\times\frac{7\left(x-1\right)}{2x}\)

=\(\frac{14}{x+1}\)

b, x=6(t/m)

khi x=6 thì A=\(\frac{14}{6+1}=2\)

c,A=7<=>\(\frac{14}{x+1}=7\)

         \(\Leftrightarrow7x+7=14\)

           \(\Leftrightarrow7x=7\Leftrightarrow x=1\left(loại\right)\)

Vậy ko có giá trị x để A=7