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Bài làm
a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(P=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x+2}\)
\(P=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{x+1}\)
\(P=\frac{x+1}{x-2}\)
b) Thay \(x=\frac{1}{2}\)vào P ta được:
\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}\)
\(P=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{2}{2}}\)
\(P=\frac{3}{2}:\frac{-1}{2}\)
\(P=\frac{3}{2}.\left(-2\right)\)
\(P=-3\)
Vậy giá trị của \(P=-3\) tại \(x=\frac{1}{2}\)
a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)
\(\Leftrightarrow P=\left(\frac{x}{x-2}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}\)
\(\Leftrightarrow P=\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{x+1}\)
\(\Leftrightarrow P=\frac{\left(x+1\right)^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{x+1}{x-2}\)
Vậy \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)
b) Ta có \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)
Thay x=\(\frac{1}{2}\left(tm\right)\)vào P ta có:
\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{4}{2}}=\frac{\frac{3}{2}}{\frac{-3}{2}}=\frac{3}{2}:\frac{-3}{2}=-1\)
Vậy \(P=-1\)khi x=\(\frac{1}{2}\)
Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
a, \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
=\(\left(\frac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}\)
\(=\frac{-3}{x-2}\)
b. Thay : x=-4
=>-3/x-2=-3/(-4)-2=1/2
câu a quy đồng mẫu lên: x^2-4=(x+2)(x-2). câu b thì thay vào. câu c toán 7 tự làm
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
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)