Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(ĐKXĐ:\hept{\begin{cases}x\ne\mp2\\x\ne3\\x\ne0\end{cases}}\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(=\left[\frac{\left(x+2\right)^2}{\left(2-x\right)\left(x+2\right)}+\frac{4x^2}{\left(2-x\right)\left(x+2\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(x+2\right)}\right]:\left[\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right]\)
\(=\frac{x^2+4x+4+4x^2-4+4x-x^2}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(=\frac{4x\left(x+2\right)}{x+2}.\frac{x}{x-3}=\frac{4x^2}{x-3}\)
1.\(A=\frac{2x^2-16x+41}{x^2-8x+22}\) \(=\frac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\frac{3}{\left(x-4\right)^2+6}\ge\frac{1}{2}\)
Dấu '' = '' xảy ra khi x = 4.
Vậy MinA= \(\frac{1}{2}\) tại x = 4.
a, điều kiện xác định là \(x\ne2;x\ne-2;x\ne0\)
\(b,\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\cdot\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=-\frac{6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)
\(=-\frac{1}{x-2}=\frac{1}{2-x}\)
c, Để A>0
mình làm hơi tắt nên chịu khó hiểu
\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)
a)\(B=\left(-\frac{\left(x+2\right)^2}{x^2-4}-\frac{4x^2}{x^2-4}+\frac{\left(x-2\right)^2}{x^2-4}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(B=\left(\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{x^2-4}\right).\frac{-x\left(x-2\right)}{\left(x-3\right)}\)
\(B=\left(\frac{-x^2-4x-4-4x^2+x-4x+4}{\left(x-2\right)\left(x+2\right)}\right).-\frac{x\left(x-2\right)}{x-3}\)
\(B=\frac{-5x^2-7x}{\left(x+2\right)}.\frac{-x}{x-3}\)
\(B=\frac{\left(-5x^2-7x\right)-x}{\left(x+2\right)\left(x-3\right)}\)
\(B=\frac{5x^3+7x^2}{\left(x+2\right)\left(x+3\right)}\)
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1
\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)
\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)
\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)
\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)
\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)
\(=\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}\)
b
Xét hơi bị nhiều TH nhá:(
Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)
TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)
\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)
Bạn tự xét nốt nhá!
c
\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)
\(\Rightarrow x=11;x=3\)
Thay vào .....
dễ mà, tik đi, mik giải cho!
Đừng mắc lừa nhưng kẻ giả vờ rủ òng thương người khác