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19 tháng 8 2020

a,\(x-\sqrt{x}-2=x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{9}{4}\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)

b, \(x\sqrt{x}+8=\sqrt{x}^3+2^3=\left(\sqrt{x}+2\right)\left(x+2\sqrt{x}+4\right)\)

c, \(x-2\sqrt{x}-3=x-2.1.\sqrt{x}+1-4\)

\(=\left(\sqrt{x}-1\right)^2-2^2=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)

d, \(x\sqrt{x}-1=\sqrt{x}^3-1^3=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

e, \(2x+3\sqrt{x}=\sqrt{x}\left(2\sqrt{x}+3\right)\)

f, \(x-7\sqrt{x}-12=\sqrt{x}^2-2.\frac{7}{2}\sqrt{x}+\frac{49}{4}-\frac{1}{4}\)

\(=\left(\sqrt{x}-\frac{7}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\)

d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)

\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)

16 tháng 7 2019

\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)

\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)

\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)

\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)

\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)

\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)

\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)

\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)

\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)

\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)

\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)

\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)

\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)

\(\text{f)}x-2\sqrt{x-1}-a^2\)

\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)

\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)

3 tháng 8 2023

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