Rút gọn: \(A=\frac{\sqrt{x^2-6x+9}}{6-2x}\)
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ĐKXĐ:
\(6-2x\ne0\)
\(x\ne3\)
\(A=\frac{\sqrt{x^2-6x+9}}{6-2x}=\frac{\sqrt{\left(x-3\right)^2}}{2.\left(3-x\right)}\)
\(=\frac{\left|x-3\right|}{2.\left(3-x\right)}\)
Với \(x-3\ge0\)thì:
\(A=\frac{x-3}{2.\left(3-x\right)}=\frac{-\left(3-x\right)}{2.\left(3-x\right)}=\frac{-1}{2}\)
Với \(x-3\le0\)thì :
\(A=\frac{3-x}{2.\left(3-x\right)}=\frac{1}{2}\)
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a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)
a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)
b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)
mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)
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a) ĐKXĐ :
\(6-2x\ne0\Leftrightarrow x\ne3\)
\(A=\frac{x^2-6x+9}{6-2x}=\frac{\left(x-3\right)^2}{-2.\left(x-3\right)}=\frac{x-3}{-2}\)
b) ĐKXĐ:
\(\sqrt{x}-7\ne0\text{ và }x\ge0\)
\(\Leftrightarrow\sqrt{x}\ne7\text{ và }x\ge0\)
\(\Leftrightarrow x\ne49\text{ và }x\ge0\)
\(B=\frac{x-49}{\sqrt{x}-7}=\frac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}=\sqrt{x}+7\)
c)
\(C=\frac{2-\sqrt{2}}{\sqrt{2}}=\frac{\left(\sqrt{2}\right)^2-\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(\sqrt{2}-1\right)}{\sqrt{2}}=\sqrt{2}-1\)
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\(1,ĐK:x\ne0;x\ne\pm6\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)
\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)
\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)
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.
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a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)