\(A=\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
\(B=\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
rút gọn biểu thức
Help me!
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right)\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
\(A=\left[\frac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{a+b+\sqrt{ab}-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\frac{a+\sqrt{ab}+b}{a-b}\right]\)
\(A=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\right].\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]\)
\(A=\frac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-\sqrt{ab}+b}\)
Điều kiện : a, b\(\ge0\)
Rút gọn:
a) \(\frac{a-b}{\sqrt{a}-\sqrt{b}}\)-\(\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)(\(a\ge0\),\(b\ge0\),\(a\ne b\))
b)\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)\(\left(a>0,b>0,a\ne b\right)\)
C)\(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)\(\left(a>0,a\ne1,a\ne4\right)\)
d)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\)\(\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)\(\left(a>0,b>0,a\ne b\right)\)
e)\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right)\):\(\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)\(\left(x>0,x\ne9\right)\)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Đề bài: Rút gọn biểu thức:
1. \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{^{ }\frac{a^4}{x^4}-1}\)
2. \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\) . \(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
3. \(\left(\frac{3}{\sqrt{1+x}}\sqrt{1-x}\right)\) : \(\left(\frac{3}{\sqrt{1-x^2}}+1\right)\)
4. \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right)\) : \(\left(\frac{a}{\sqrt{ab+b}}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
5. \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}\) + \(\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\) .\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
Các bạn giúp tớ nhé, hứa sẽ tick, tớ cảm ơn!!!!
1.
Đặt \(\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}\)
\(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}\)
\(=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}\)
\(=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}\)
\(=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}\)
2.
\(=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})\)
\(=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)=(\sqrt{a}+1)^2(\sqrt{a}-1)^2\)
\(=(a-1)^2\)
3.
\(=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}\)
4. Bạn xem lại đề xem đã đúng chưa?
5.
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}\)
Giúp mình nhé, mình đang càn gấp :<<<
a) \(\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)\)
b) \(\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\)
c) \(\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)\)
d) \(\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)\)
\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)
\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)
\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
RÚT GỌN CÁC BIỂU THỨC SAU
\(A=\frac{-2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(B=\left(\sqrt{ab}+2\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}+\sqrt{\frac{1}{ab}}}\right).\sqrt{ab}\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)