\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
a, Rút gọn biểu thức
b,Tìm x để A\(\le\)-2
mn giúp e vs ạ
K
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24 tháng 7 2021
\(ĐKXĐ:x\ne2;x\ne-2;x\ne0\)
\(a,P=\left(\frac{-1}{2-x}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{-2-x+2-x-2x}{\left(2-x\right)\left(2+x\right)}\right)\left(\frac{2-x}{x}\right)\)
\(P=\frac{-4x}{\left(2-x\right)\left(2+x\right)}\frac{2-x}{x}\)
\(P=\frac{-4}{2+x}\)
\(b,P=\frac{-4}{2+x}=\frac{1}{2}\)
\(2+x=-8\)
\(x=-10\)
\(c,P=-\frac{4}{2+x}\)
\(< =>-4⋮x+2\)
lập bảng ra thì bạn ra đc \(x=\left\{2;-1;-3;-6\right\}\)
AD
24 tháng 7 2021
a)\(P=\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right)\left(\frac{2}{x}-1\right)\)
\(P=\left(\frac{1}{x-2}+\frac{2x}{\left(x+2\right)\left(x-2\right)}+\frac{1}{2+x}\right).\frac{2-x}{x}\)
\(P=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)
\(P=\frac{-4}{x+2}\)
b) Để P=1/2
\(\Rightarrow-\frac{4}{x+2}=\frac{1}{2}\)
\(\Leftrightarrow-8=x+2\)
\(\Leftrightarrow x=-10\)
c) Để P nhận GT nguyên
\(\Rightarrow\left(x+2\right)\inƯ_{\left(-4\right)}\)
\(\Rightarrow\left(x+2\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow x=\left\{-3;-1;-4;0;-6;2\right\}\)
#H
2 tháng 10 2020
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)
\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)
b) \(18A=1\)
<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))
<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)
<=> 18( x2 - 4x + 5 ) = x3 + 6x2 - 32
<=> 18x2 - 72x + 90 = x3 + 6x2 - 32
<=> x3 + 6x2 - 32 - 18x2 + 72x - 90 = 0
<=> x3 - 12x2 + 72x - 122 = 0
Rồi đến đây chịu á :)
KN
11 tháng 2 2020
a) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\frac{2x+4}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{x^2-4}=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{\left(x+2\right)}=\frac{-2x+4}{x+2}\)
b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
+) x = 0 \(\Rightarrow A=\frac{-2.0+4}{0+2}=\frac{4}{2}=2\)
+) x = 3 \(\Rightarrow A=\frac{-2.3+4}{3+2}=\frac{-2}{5}\)
30 tháng 3 2021
a) ĐKXĐ : x ≠ ±2
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\div\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=\frac{-1}{x-2}\)
b) Để A < 0 thì -1/x-2 < 0
=> x - 2 > 0 <=> x > 2
Vậy với x > 2 thì A < 0
29 tháng 12 2017
\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(E=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)^2}\right).\left(\frac{\left(1-x\right)\left(1+x\right)}{2}\right)^2\)
\(E=\left(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)^2}-\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\frac{\left(1-x\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{\left(x-2\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{2\left(x-2\right)\left(x-1\right)}{4}\)
\(E=\frac{\left(x-2\right)\left(x-1\right)}{2}\)
29 tháng 12 2017
a) \(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x+2}{\left(x+1\right)^2}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{x^2-3x+2-x^2-3x-2}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\frac{-6x.\left(x^2-1\right)^2}{\left(x-1\right)^2\left(x+1\right).4}=\frac{-3x\left(x^2-1\right)^2}{\left(x^2-1\right)\left(x-1\right).4}=\frac{-3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right).4}\)\(=\frac{-3x\left(x+1\right)}{4}\)
b) Muốn \(\frac{E-4}{5}=x\) thì \(\frac{\frac{-3x\left(x+1\right)}{4}-4}{5}=x\)
\(\Rightarrow\frac{\frac{-3x^2\left(x+1\right)}{4}-\frac{16}{4}}{5}=x\)
\(\Rightarrow\frac{-3x^3-3x^2-16}{4}=5x\)
\(\Rightarrow-3x^3-3x^2-16=20x\)
\(\Rightarrow-3x^3-3x^2-16=20x\).....................................................................
Đề sai ! Sửa nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow A=\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x-2}\right)\)
\(\Leftrightarrow A=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow A=\frac{2x+4-4}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(\Leftrightarrow A=\frac{2x\left(x-2\right)}{-x\left(x+2\right)}\)
\(\Leftrightarrow A=-\frac{2\left(x-2\right)}{x+2}\)
b) Để \(A\le-2\)
\(\Leftrightarrow-\frac{2\left(x-2\right)}{x+2}\le-2\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{x+2}\ge2\)
\(\Leftrightarrow\frac{x-2}{x+2}\ge1\)
\(\Leftrightarrow x-2\ge x+2\)
\(\Leftrightarrow-2\ge2\)(ktm)
Vậy để \(A\le-2\Leftrightarrow x\in\varnothing\)
a.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)
\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)
\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)