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Lời giải:
a.
$7-3a=(\sqrt{7}-\sqrt{3a})(\sqrt{7}+\sqrt{3a})$
b.
$14x^2-11=(\sqrt{14}x-\sqrt{11})(\sqrt{14}x+\sqrt{11})$
c.
$3x-6\sqrt{x}-6=3(x-2\sqrt{x}-2)$
$=3[(\sqrt{x}-1)^2-3]$
$=3(\sqrt{x}-1-\sqrt{3})(\sqrt{x}-1+\sqrt{3})$
d.
$x\sqrt{x}-3\sqrt{x}-2=x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2$
$=x(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}-2)+(\sqrt{x}-2)$
$=(\sqrt{x}-2)(x+2\sqrt{x}+1)$
$=(\sqrt{x}-2)(\sqrt{x}+1)^2$
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
a/ \(x^2+5\sqrt{x}+6=x^2+2\sqrt{x}+3\sqrt{x}+6\)
\(=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
b/ \(x^2+4\sqrt{x}+3=x^2+\sqrt{x}+3\sqrt{x}+3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)
c/ k bik làm
a) \(x-4\sqrt{x-2}+2\left(x\ge2\right)\)
\(=x-4\sqrt{x-2}-2+4\)
\(=\left(x-2\right)-4\sqrt{x-2}+4\)
\(=\left(\sqrt{x-2}\right)^2-2\cdot2\cdot\sqrt{x-2}+2^2\)
\(=\left(\sqrt{x-2}-2\right)^2\)
b) \(x+4\sqrt{x-2}+2\left(x\ge2\right)\)
\(=x+4\sqrt{x-2}+4-2\)
\(=\left(x-2\right)+4\sqrt{x-2}+4\)
\(=\left(\sqrt{x-2}\right)^2+2\cdot2\cdot\sqrt{x-2}+2^2\)
\(=\left(\sqrt{x-2}+2\right)^2\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
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) \(x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)\)
\(=x\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)
b) \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
a, \(x-\sqrt{x}\)= \(\sqrt{x}.\left(\sqrt{x}-1\right)\)
b, 3x+6\(\sqrt{x}\)= \(\sqrt{x}.\left(3\sqrt{x}+6\right)\)
c, x+2\(\sqrt{x}+1\)= \(\left(\sqrt{x}\right)^2+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)
=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)
=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)