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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
\(1,ĐK:x\ne0;x\ne\pm6\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)
\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)
\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)
Cho tam giác ABC vuông tại B có góc B1=B2 ; Â=60o, kẻ BH vuông góc với AC (H thuộc AC). Qua B kẻ đường thẳng d song song với AC.
a) Tính góc ABH.
b) Chứng minh đường thẳng d vuông góc với BH.
a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)
a: \(P=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)\cdot\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\cdot\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x^3-x-2x+2}{x^2+x+1}\right)\)
\(=\left(\dfrac{x}{x+2}-\dfrac{x^2-2x+4}{\left(x+2\right)^2}\right):\left(\dfrac{1}{x+2}\cdot\dfrac{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{x^2+2x-x^2+2x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x^2+x-2\right)}{x^2+x+1}\right)\)
\(=\dfrac{4x-4}{\left(x+2\right)^2}:\left(\dfrac{1}{x+2}\cdot\dfrac{\left(x-1\right)\left(x+2\right)\left(x-1\right)}{x^2+x+1}\right)\)
\(=\dfrac{4\left(x-1\right)}{\left(x+2\right)^2}\cdot\dfrac{x^2+x+1}{\left(x-1\right)^2}=\dfrac{4\left(x^2+x+1\right)}{\left(x+2\right)^2\left(x-1\right)}\)
b: Để P>0 thì x-1>0
hay x>1
ĐKXĐ: \(x\ne\pm2\)
a)\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+4}{x^2-4}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+4}{x^2-4}\)
\(=\frac{x+2}{x^2-4}+\frac{x-2}{x^2-4}+\frac{x^2+4}{x^2-4}=\frac{x+2+x-2+x^2+4}{x^2-4}=\frac{x^2+2x+4}{x^2-4}=\frac{\left(x+1\right)^2+3}{x^2-4}\)
b) \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+3\ge3>0\)
=> A<0 khi \(x^2-4< 0\Leftrightarrow x^2< 4\)
Vì \(x^2\ge0\Rightarrow0\le x^2< 4\Leftrightarrow-2< x< 2\)
Tại sao lại x khác -1 thì A<0 vì khi x=-1 thì A=-1<0 mà!