Chứng minh các đẳng thức sau :
a) \(\left(\dfrac{1-a\sqrt{a}}{a-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0;a\ne1\)
b) \(\dfrac{a+b}{b^2}\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\left|a\right|\) với \(a+b>0;b\ne0\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
LH
Lưu Hạ Vy
22 tháng 4 2017
Đúng(0)
Những câu hỏi liên quan
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: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
30 tháng 7 2022
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
28 tháng 10 2021
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-1}{1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
BN
4 tháng 7 2023
a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm
b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm
28 tháng 10 2021
a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
8 tháng 12 2023
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
AT
15 tháng 7 2021
a) \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{\left(2+\sqrt{a}-\sqrt{a}-1\right)\left(2+\sqrt{a}+\sqrt{a}+1\right)}{2\sqrt{a}+3}\)
\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)
b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
15 tháng 7 2021
a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)
\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)
Vậy ta có đpcm
b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)
Vậy ta có đpcm
N
19 tháng 11 2021
\(a,VT=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{\left(x+1\right)\left(1-3x\right)}{3x}\right)\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2-6x}{3x}\right)\cdot\dfrac{x}{x-1}=\dfrac{6x}{3x}\cdot\dfrac{x}{x-1}=\dfrac{2}{x-1}=VP\left(x\ne0;x\ne1\right)\)
\(b,VT=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}=VP\left(a\ge0;a\ne1\right)\)
