Rút gọn biểu thức
\(\left(a+b-\frac{2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
Moinj người giúp em sớm nhất có thể với ạ ? Em cảm ơn :)
Bai 1)A=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right),\)
a)rút gọn A
2) B=1+\(\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}-1}{1-a\sqrt{a}}\right).\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right),\)
a) rút gọn B
b) tìm a để B=\(\frac{\sqrt{6}}{1+\sqrt{6}}\)
c) chứng minh rằng B>2/3
Giúp mk với nếu đúng mk tick cho nha ,cảm ơn
\(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(2a+2\sqrt{ab}+2b\right)}
\)
a. Rút gọn P
b. Tìm giá trị nguyên của a để giá trị P nguyên
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
Tìm ĐKXĐ và rút gọn các biểu thức sau:
\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}\)
\(B=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}.\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(B=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\)\(\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
Giúp mình giải bài toán phía dưới với nha mọi người
MÌNH CẢM ƠN RẤT NHIỀU!
Cho biểu thức
A=\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}1}\right)\)
a)Rút gọn A
b) Tìm giá trị của x để A >-6
#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
#)Giải :
b) Để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Kết hợp với đkxđ => 0 < x < 9
Rút gọn biểu thức
\(\frac{\frac{\left(a-b\right)3}{\left(\sqrt{a}-\sqrt{b}\right)3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\)
Cho biểu thức ;
\(P=\frac{a}{\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{b}\right)}-\frac{b}{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{a}\right)}-\frac{ab}{\left(1+\sqrt{a}\right)\left(1-\sqrt{b}\right)}\)
a, rút gọn biểu thức
b tìm a, b là các số chính phương để P=2
cần gấp ạ , thanks mn
a/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}\)
b/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}=2\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{b}+1\right)=1\)
Xong rồi nhá
Giúp mình rút gọn biểu thức sau đi :<<
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\)
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\) \(ĐKXĐ:\hept{\begin{cases}a\ge0\\b\ge0\\a\ne b\end{cases}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\right)\left(\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)
\(=1\)
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\left(\sqrt{a}+\sqrt{b}\right)-b\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\cdot\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)^2}.\)\(=1\)
1) Rút gọn biểu thức sau :
M= \(\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{âb}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\)
với b>a>0
giúp mk với
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)