Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(A=\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right)\left(\frac{2}{x-1}\right)\)ĐK : \(x\ne1;\pm2\)
\(=\left(\frac{x+2+2x+x-2}{x^2-4}\right)\left(\frac{2}{x-1}\right)=\frac{4x}{x^2-4}.\frac{2}{x-1}=\frac{8x}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)
b, bạn check lại đề bài nhé
A=(1x−2 +2xx2−4 +1x+2 )(2x−1 )ĐK : x≠1;±2
=(x+2+2x+x−2x2−4 )(2x−1 )=4xx2−4 .2x−1 =8x(x−1)(x−2)(x+2)
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
a) -4x2+2x
b) -4x2+2x=0
x(-4x+2)=0
=> x=0 hoặc -4x+2=0
-4x = -2
x=1/2(đpcm)
c) Thay x=-1/4 vào -4x2+2x ta có : -4 (-1/4)2 +2(-1/4) = ... (tự tính )
a) A = (x - 3)(x + 1) - (2x - 1)^2 - (x + 2)(x - 2)
A = x^2 - 2x - 3 - 4x^2 + 4x - 1 - x^2 + 4
A = -4x^2 + 2x
b) 4x^2 - 2x = 0
<=> 2x(2x - 1) = 0
<=> 2x = 0 hoặc 2x - 1 = 0
<=> x = 0 hoặc x = 1/2
c) với x = -1/4, ta có:
4(-1/4)^2 - 2(-1/4) = 3/4
a) \(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(A=4x^2+4x+1-x^2+4-2x^2-2x\)
\(A=x^2+2x+5\)
b) Để A = 4
=> \(x^2+2x+5=4\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c) Ta có A = x2 + 2x + 5
A = ( x + 1 )2 + 4
=> \(A\ge4>0\left(đpcm\right)\)
a,\(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(=4x^2+4x+1-x^2+4-2x^2-2x\)
\(=x^2+2x+5\)
b,\(A=x^2+2x+5=4\)
\(\Rightarrow x^2+2x+5-4=0\)
\(x^2+2x+1=0\)
\(\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
c, Ta có: \(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4>0\)
Hay: A > 0 => đpcm
=.= hok tốt!!
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
a, \(A=\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right).\left(\frac{2}{x}-1\right)\) \(\left(ĐK:x\ne\pm2\right)\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\left(\frac{2}{x}-\frac{x}{x}\right)\)
\(A=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(A=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{-\left(x-2\right)}{x}\)
\(A=\frac{-4x.\left(x-2\right)}{\left(x-2\right)\left(x+2\right).x}\)
\(A=\frac{-4}{x+2}\)
b, \(A=\frac{-4}{x+2}=1\)
\(\rightarrow\frac{-4}{x+2}=\frac{x+2}{x+2}\)
\(\rightarrow-4=x+2\)
\(\rightarrow-6=x\)
a) ĐKXĐ của A là x\(\ne\pm2\); x\(\ne1\)
Ta có
A= \((\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2})\cdot\frac{2}{x-1}\)
A=\(\frac{x+2+2x+x-2}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\frac{2}{x-1}\)
A=\(\frac{6x}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)