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)

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 có nghĩa khi \(\hept{2x-2\ne02-2x^2\ne0\Leftrightarrow\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}\Leftrightarrow}x\ne\pm1}\)

Vậy A có nghĩa khi \(x\ne\pm1\)

b) \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\left(x\ne\pm1\right)\)

\(\Leftrightarrow A=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)

\(\Leftrightarrow\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow A=\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)

Vậy A=\(\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)

b) \(A=\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)

A=\(\frac{-1}{2}\)\(\Leftrightarrow\frac{1}{2\left(x-1\right)}=\frac{-1}{2}\)

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

<=> x-1=-1

<=> x=0 (tmđk)

Vậy x=0 thì \(A=\frac{-1}{2}\)

a) \(x\ne1,2;x\inℝ\)

a) Để biểu thức A có nghĩa \(\hept{\begin{cases}2-2x^2\ne0\\2x-2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}2\left(1-x\right)\left(1+x\right)\ne0\\2\left(x-1\right)\ne0\end{cases}}\Rightarrow x\ne\pm1\)

b) Rút gọn \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

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

c) \(A=-\frac{1}{2}\Leftrightarrow\frac{1}{2\left(x+1\right)}=-\frac{1}{2}\Leftrightarrow\frac{1}{x+1}=-1\Leftrightarrow x+1=-1\Leftrightarrow x=-2\)Vậy x=-2 Thì A=-1/2

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

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 \(x\in\left\{-2;0;-3;1\right\}\)

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

b) Ta có:  x - 3 = 10 => x = 13

Ta cố bdt \(|a|+|b|\ge|a+b|\), dễ dàng chứng mình bằng bình phương 2 vế. Dấu = sảy ra <=>IaI.IbI=a.b <=> a.b>=0

áp dụng vào từng câu

a)A=Ix+1I+Ix+2I+Ix+3I+I-x-4I+I-x-5I  ( vì Ix+4I=I-x=4I, Ix+5I=I-x-5I

A>=I(x+1)+(-x-5)I+I(x+2)+(-x-4)I +Ix+3I=4+2+Ix+3I=6+Ix+3I>=6

Dấu bằng khi (x+1)(-x-5)>=0;(x+2)(-x-4)>=0;Ix+3I=0 =>x=-3

b) LÀm tương tự MinB=18

Dấu = khi (2x+1)(-2x-11)>=0;(2x+3)(-2x-9)>=0;(2x+5)(-2x-7)>=0 <=>-7/2<=x<=-5/2