K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

13 tháng 9 2019

ĐK: \(x\ge-7\)

PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)

\(\Leftrightarrow x=9\) 

P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((

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

\(b)\) \(B=a-b=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)\(\Rightarrow\)\(B^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)

\(B^2=4-2\sqrt{4-3}=4-2=2\)\(\Rightarrow\)\(B=\sqrt{2}\) ( vì \(B>0\) ) 

... 

19 tháng 12 2018

cảm ơn nhe <3 :)) 

8 tháng 10 2019

2.\(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}=6-4\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}+4\sqrt{x}=6\)

\(\Leftrightarrow2\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=\frac{6}{2}\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2=\left(3\right)^2\)

\(\Leftrightarrow x=9\)

vậy x=9 

mình chỉ giúp bạn được vậy thui :)

chúc bạn học tốt nha:)))

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) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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