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\(a,\dfrac{x^2}{x-3}-\dfrac{6x}{x-3}+\dfrac{9}{x-3}\\ =\dfrac{x^2-6x+9}{x-3}\\ =\dfrac{\left(x-3\right)^2}{x-3}=x-3\)
a: \(=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)
b: \(=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2x+2}\)
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\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)
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a: Xét tứ giác ABDC có
N là trung điểm của BC
N là trung điểm của AD
Do đó: ABDC là hình bình hành
mà \(\widehat{BAC}=90^0\)
nên ABDC là hình chữ nhật
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a: Xét ΔAEB vuông tại E và ΔAFC vuông tại F có
\(\widehat{BAC}\) chung
Do đó:ΔAEB\(\sim\)ΔAFC
Suy ra: AE/AF=AB/AC
hay \(AE\cdot AC=AB\cdot AF\)
b: Xét ΔHFB vuông tại F và ΔHEC vuông tại E có
\(\widehat{FHB}=\widehat{EHC}\)
Do đó:ΔFHB\(\sim\)ΔEHC
Suy ra: HF/HE=HB/HC
hay \(HF\cdot HC=HB\cdot HE\)
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1.\(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=4\)
\(\Leftrightarrow8x=4\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
Vậy S = \(\dfrac{1}{2}\)
2.\(ĐK:x\ne1;-3\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}=-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+3x+x+3-x^2+x-2x+2=-4\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\left(ktm\right)\)
Vậy S vô nghiệm
1) ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+4x+4-x^2+4x-4=4\)
\(\Leftrightarrow8x=4\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
Vậy ....
2) ĐKXĐ:\(x\ne1,-3\)
\(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\)
\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{4}{x^2-x+3x-3}=0\)
\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Rightarrow x^2+4x+3-x^2-x+2+4=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\)(ktm)