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29 tháng 6 2019

\(a,\)\(7\sqrt{ab}+7b-\sqrt{a}-\sqrt{b}\)

\(=7\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(7\sqrt{b}-1\right)\)

\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)

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

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

\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)

\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)

\(=\sqrt{x-5y}\left(\sqrt{x-5y}-1\right)\)

a, \(7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}=7\sqrt{B}\left(\sqrt{A}+\sqrt{B}\right)-\left(\sqrt{A}+\sqrt{B}\right)\)\(=\left(\sqrt{A}+\sqrt{B}\right)\left(7\sqrt{B}-1\right)\)

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

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

c,\(\sqrt{x^2-25y^2}-\sqrt{x-5y}=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)

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

29 tháng 6 2019

\(a,7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}\)(  Với A>= 0,  B>=0)

\(=\left(7\sqrt{AB}-\sqrt{A}\right)+\left(7B-\sqrt{B}\right)\)

\(=7\sqrt{A}\left(\sqrt{B}-1\right)+7\sqrt{B}\left(\sqrt{B}-1\right)\)

\(=\left(\sqrt{B}-1\right)\left(7\sqrt{A}+7\sqrt{B}\right)\)

\(=7\left(\sqrt{B}-1\right)\left(\sqrt{A}+\sqrt{B}\right)\)

29 tháng 6 2019

\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)Với a>= 0,  b>=0)

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

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

\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)

\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)

\(=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)

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

AH
Akai Haruma
Giáo viên
10 tháng 8 2021

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.

$x\sqrt{x}-3\sqrt{x}-2=x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2$
$=x(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}-2)+(\sqrt{x}-2)$

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

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

8 tháng 7 2017

a) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

\(=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

b) \(7\sqrt{ab}+7b-\sqrt{a}-\sqrt{b}\)

\(=7\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{a}+\sqrt{b}\right)\left(7\sqrt{b}-1\right)\)

c) \(a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)

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

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

d) \(\sqrt{x^2-25y^2}-\sqrt{x-5y}\)

\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)

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

4 tháng 10 2020

a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)

\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)

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

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

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

4 tháng 10 2020

b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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

13 tháng 11 2021

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

13 tháng 11 2021

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

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