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a, Rút gọn :
\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{1\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{x-5+2x+10-2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{x+15}{\left(x+5\right)\left(x-5\right)}\)
Đức Hiệp Tùng Giúp tôi giải toán
3 phút trước (13:18)
Kb đi buồn quá
a) ĐKXĐ: \(x\ne2\); x \(\ne\)-2
Ta có: P = \(\left(\frac{2+x}{x-2}+\frac{2}{x+3}-\frac{x^2+5x}{x^2-4}\right):\left(1-\frac{x+1}{x+2}\right)\)
P = \(\left(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2+5x}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x-1}{x+2}\right)\)
P = \(\left(\frac{x^2+4x+4+2x-4-x^2-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{1}{x+2}\)
P = \(\frac{x}{\left(x-2\right)\left(x+2\right)}\cdot\left(x+2\right)\)
P = \(\frac{x}{x-2}\) (đk: x khác 2)
b) Ta có: x2 - 2x = 0
=> x(x - 2) = 0
=> \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\left(tm\right)\\x=2\end{cases}}\)
Vì biểu thức P x \(\ne\)2 => x = 0=> P = \(\frac{0}{0-2}=0\)
cái này nó hơi khó 1 tí nên chú ý chút khác lên lever :>
a, \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)ĐK : x khác 0 ; 2 ; -2
\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4x\left(x-2\right)}{MTC}+\frac{2x\left(x+2\right)}{MTC}+\frac{\left(6-5x\right)x}{MTC}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4x^2-8x+2x^2+4x+6x-5x^2}{MTC}\right):\frac{x+1}{x-2}\)
\(=\frac{x^2+2x}{x\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : \(x^2-2x=8\Leftrightarrow x^2-2x-8=0\)
\(\left(x-4\right)\left(x+2\right)=0\)<=> \(x=4;-2\)
TH1 : Thay x = 4 ta được : \(\frac{1}{4+1}=\frac{1}{5}\)
TH2 : Thay x = -2 ta được : ( ktmđkxđ )
\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)
a)\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\right)\times\frac{x-2}{x+1}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)
\(=\left(\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\times\frac{x-2}{x+1}\)
\(=\frac{1}{x+1}\)
b) x2 - 2x = 8
<=> x2 - 2x - 8 = 0
<=> x2 - 4x + 2x - 8 = 0
<=> x( x - 4 ) + 2( x - 4 ) = 0
<=> ( x - 4 )( x + 2 ) = 0
<=> x = 4 ( tm ) hoặc x = -2 ( ktm )
Với x = 4 ( tm ) => A = 1/5
Với x = -2 ( ktm ) => A không xác định
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
a,\(ĐKXĐ:\hept{\begin{cases}x\ne\mp2\\x\ne3\\x\ne0\end{cases}}\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(=\left[\frac{\left(x+2\right)^2}{\left(2-x\right)\left(x+2\right)}+\frac{4x^2}{\left(2-x\right)\left(x+2\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(x+2\right)}\right]:\left[\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right]\)
\(=\frac{x^2+4x+4+4x^2-4+4x-x^2}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(=\frac{4x\left(x+2\right)}{x+2}.\frac{x}{x-3}=\frac{4x^2}{x-3}\)
ĐKXĐ: x\(\ne\)1, x\(\ne\)-1
MTC (x-1)(x+1)
\(\Leftrightarrow\)(\(\frac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)+ \(\frac{2\left(x-1\right)}{MTC}\)-\(\frac{-\left(5-x\right)}{MTC}\)) : \(\frac{1-2x}{MTC}\)
\(\Rightarrow\)\(\left[-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)\right]:\left(1-2x\right)\)
\(\Leftrightarrow\frac{-x-1+2x-2+5-x}{1-2x}\)
=\(\frac{-2x+2x+2}{1-2x}\)
=\(\frac{2}{1-2x}\)
b. mình chỉ biết \(x< \frac{1}{2}\) thôi chứ ko biết làm sao
hình như là giải Bất phương trình \(\frac{2}{1-2x}>0\)
\(A=\left(\frac{x^2-16}{x-4}+1\right):\left(\frac{x-2}{x-3}+\frac{x+3}{x+1}+\frac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+5\right):\left(\frac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}+\frac{x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2+x-2x-2+x^2-9+x+2-x^2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x^2-9}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+5\right):\left(\frac{x+3}{x+1}\right)=\frac{x+3}{\left(x+5\right)\left(x+1\right)}\)