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\(x=3+2\sqrt{2}\)
\(x-3-2\sqrt{2}=0\)
\(x-\left(3+2\sqrt{2}\right)=0\) Vậy nhân tử của \(x=3+2\sqrt{2}\) là \(x-\left(3+2\sqrt{2}\right)\)
2
\(M=2y-3x\sqrt{y}+x^2=y-2x\sqrt{y}+x^2+y-x\sqrt{y}\\ =\left(\sqrt{y}-x\right)^2+\sqrt{y}\left(\sqrt{y}-x\right)\\ =\left(\sqrt{y}-x\right)\left(\sqrt{y}-x+\sqrt{y}\right)\\ =\left(\sqrt{y}-x\right)\left(2\sqrt{y}-x\right)\)
b
\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)
Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:
\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)
1:
a: =>2x-2căn x+3căn x-3-5=2x-4
=>căn x-8=-4
=>căn x=4
=>x=16
b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>(căn x-2)(x-căn x+4)=0
=>căn x-2=0
=>x=4
a) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
b) \(x-\sqrt{x}-2=\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+1\right)\)
c) \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
a) \(x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)\)
\(=x\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)
b) \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
\(=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(x^3+\left(2m+5\right)x^2+\left(2m+6\right)x-4m-12=\left(x^3-x^2\right)+\left[\left(2m+6\right)x^2-\left(2m+6\right)x\right]+\left[\left(4m+12\right)x-\left(4m+12\right)\right]=\left[x^2+\left(2m+6\right)x+\left(4m+12\right)\right]\left(x-1\right)\)
\(12-\sqrt{x}-x=\left(4-\sqrt{x}\right)\left(3+\sqrt{x}\right)\)