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\(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\div\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\dfrac{x-1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4x\sqrt{x}-4\sqrt{x}}{\left(x-1\right)}\times\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
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\(\dfrac{4x^2}{\left(x-1\right)^2}=2\)
\(\Leftrightarrow4x^2=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow2\left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.\) (nhận)
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\(x=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(16-15\right)\left(4+\sqrt{15}\right)}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
= 5 - 3 = 2
\(M=\dfrac{4x^2}{\left(x-1\right)^2}=16\)
a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)
c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)
d, câu này sai đề rồi , nếu sửa lại phải như này :
\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)
= \(-5x\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))
= \(2x-8\)
a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)
Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)
\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))
= \(2x-8\)
a) Ta có: \(A=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-2\cdot2\cdot\sqrt{3}+3}-\sqrt{4+2\cdot2\cdot\sqrt{3}\cdot3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-\left(2+\sqrt{3}\right)\)(Vì \(2>\sqrt{3}>0\))
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
b) Ta có: \(B=\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}\right)\cdot\frac{\left(\sqrt{x}+2\right)\cdot\left(x-4\right)}{\sqrt{x}}\)
\(=\frac{x+3\sqrt{x}+2-\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}\cdot\frac{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\frac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{6\sqrt{x}}{\sqrt{x}}=6\)
a) Bình phương lên ta đc
\(A^2=7-4\sqrt{3}+7+4\sqrt{3}-2\sqrt{7^2-\left(4\sqrt{3}\right)^2}=14-2=12\)
\(\Rightarrow A=\mp\sqrt{12}\)
a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b: ĐểP<15/4 thì P-15/4<0
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)
=>12 căn +32-15 căn x+30<0
=>-3 căn x<-62
=>căn x>62/3
=>x>3844/9
Có: \(\sqrt{x+4\sqrt{x-4}}\)=\(\sqrt{x-4+2\sqrt{x-4}+4}\)=\(\sqrt{\left(\sqrt{x-4}+2\right)^2}\)=\(\sqrt{x-4}\)+2
\(\sqrt{x-4\sqrt{x-4}}\)=\(\sqrt{x-4-4\sqrt{x-4}+4}\)=\(\sqrt{\left(\sqrt{x-4}-2\right)^2}\)=2-\(\sqrt{x-4}\) vì x<8
suy ra x-4<4 \(\Rightarrow\)\(\sqrt{x-4}\)<2\(\Rightarrow\)\(\sqrt{x-4}\)-2<0
Khi đó biểu thức Q trở thành
Q=\(\sqrt{x-4}\)+2+2-\(\sqrt{x-4}\)=4
vậy Q=4