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a/ \(A=\frac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}\)
b/ Thay x = 25 vào A ta được:
\(A=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-2}=\frac{5}{3}\)
c/ A = -1/3 \(\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=-\frac{1}{3}\Rightarrow2-\sqrt{x}=3\sqrt{x}\)
\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
Vậy x = 1/4
a) \(ĐKXĐ:x\ge0;x\ne3\)
b) \(A=\left(\frac{x-2\sqrt{3x}+3}{x-3}\right)\left(\sqrt{4x}+\sqrt{12}\right)\)
\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-\sqrt{3}\right)^2}{\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)}\right)\left(2\sqrt{x}+2\sqrt{3}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x}+\sqrt{3}}\right).2\left(\sqrt{x}+\sqrt{3}\right)\)
\(\Leftrightarrow A=2\left(\sqrt{x}-\sqrt{3}\right)\)
\(\Leftrightarrow A=2\sqrt{x}-2\sqrt{3}\)
c) Thay \(x=4-2\sqrt{3}\)vào A, ta có :
\(A=2\sqrt{4-2\sqrt{3}}-2\sqrt{3}\)
\(\Leftrightarrow A=2\sqrt{\left(1-\sqrt{3}\right)^2}-2\sqrt{3}\)
\(\Leftrightarrow A=2\left(\sqrt{3}-1\right)-2\sqrt{3}\)
\(\Leftrightarrow A=2\sqrt{3}-2-2\sqrt{3}\)
\(\Leftrightarrow A=-2\)
\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)
\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)
b)\(\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=1\)
\(TH1:x-3>=0\)
\(< =>x+3>=0\)
\(\left|x-3\right|-\left|x+3\right|=1\)
\(x-3-x-3=1\)
\(-6=1\)(loại)
\(TH2:x-3< =0\)
\(x+3>=0\)
\(< =>\left|x-3\right|-\left|x+3\right|=1\)
\(3-x-x-3\)
\(-2x=1\)
\(x=-\frac{1}{2}\left(TM\right)\)
\(TH3:x-3< =0\)
\(x+3< =0\)
\(< =>\left|x-3\right|-\left|x+3\right|=1\)
\(3-x+X+3=1\)
\(6=1\)(loại)
\(< =>x=\left\{\frac{1}{2}\right\}\)để \(A=1\)
a) \(A=5x-\sqrt{4x^2-4x+1}\)
\(=5x-\sqrt{\left(2x-1\right)^2}\)
\(=5x-\left|2x-1\right|\)
+) Với x < 1/2
A = 5x - [ -( 2x - 1 ) ] = 5x - ( 1 - 2x ) = 5x - 1 + 2x = 7x - 1
+) Với x ≥ 1/2
A = 5x - ( 2x - 1 ) = 5x - 2x + 1 = 3x + 1
b) Với x = -2 < 1/2
=> A = 7.(-2) - 1 = -14 - 1 = -15
ĐKXĐ : \(x\in R\)
a) \(A=6x-1+\sqrt{x^2-4x+4}\)
\(A=6x-1+\sqrt{\left(x-2\right)^2}\)
\(A=6x-1+\left|x-2\right|\)
b) Khi \(x=5\)ta có \(A=6\cdot5-1+\left|5-2\right|=32\)
c) \(A=1\)
\(\Leftrightarrow6x-1+\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|=2-6x\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow x-2=2-6x\)
\(\Leftrightarrow7x=4\)
\(\Leftrightarrow x=\frac{4}{7}\)( không thỏa )
+) Xét \(x< 2\)
\(pt\Leftrightarrow x-2=6x-2\)
\(\Leftrightarrow-5x=0\)
\(\Leftrightarrow x=0\)( thỏa )
Vậy....