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mik viết nhầm đầu bài
khi x=4+2√3 thì giá trị của p=√x -2/3√x bằng bao nhiêu
\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\Leftrightarrow A^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}\)
\(A^2=6+4=10\Rightarrow A=\sqrt{10}\)
\(3\sqrt{8}-\sqrt{50}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=6\sqrt{2}-5\sqrt{2}-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}-\sqrt{2}+1\)
\(=1\)
Phép 1:
Ta có: \(3\cdot\sqrt{7-4\sqrt{3}}\)
\(=3\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=3\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
Phép 2:
Ta có: \(\sqrt{11+4\sqrt{7}}\)
\(=\sqrt{7+2\cdot\sqrt{7}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{7}+2\right)^2}\)
\(=\left|\sqrt{7}+2\right|\)
\(=\sqrt{7}+2\)(Vì \(\sqrt{7}+2>0\))
Phép 3:
Ta có: \(2\cdot\sqrt{11-4\sqrt{7}}\)
\(=2\cdot\sqrt{7-2\cdot\sqrt{7}\cdot2+4}\)
\(=2\cdot\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=2\cdot\left|\sqrt{7}-2\right|\)
\(=2\cdot\left(\sqrt{7}-2\right)\)(Vì \(\sqrt{7}>2\))
\(=2\sqrt{7}-4\)
Phép 4:
Ta có: \(\sqrt{19-4\sqrt{15}}\)
\(=\sqrt{15-2\cdot\sqrt{15}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{15}-2\right)^2}\)
\(=\left|\sqrt{15}-2\right|\)
\(=\sqrt{15}-2\)(Vì \(\sqrt{15}>2\))
a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=9/4 thì A=3/2:1/2=3/2*2=3
a) Ta có: \(\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}-\sqrt{3}\)
c) Ta có: \(\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{6-2\cdot\sqrt{6}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|\sqrt{6}-\sqrt{5}\right|\)
\(=\sqrt{6}-\sqrt{5}\)
d) Ta có: \(\sqrt{13-4\sqrt{3}}\)
\(=\sqrt{12-2\cdot\sqrt{12}\cdot1+1}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=\left|2\sqrt{3}-1\right|\)
\(=2\sqrt{3}-1\)
g) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)