Chứng minh các đẳng thức sau:
a) \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0;a\ne1\)
b) \(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}=\left|a\right|\) với \(a+b>0;b\ne0\)
Chứng minh các đẳng thức sau:
a) \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)
(Với \(x\ge0;x\ne1\))
b) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}-b}=2\sqrt{a}\)
(Với a>0; b>0; \(a\ne b\))
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\left(a\ge0,a\ne1\right)\)
Chứng minh đẳng thức trên
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(ĐPCM\right)\)
Chứng minh các đẳng thức sau:
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0\)và \(a\ne0\)
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)
Biến đổi vế trái ta có:
\(=\left[\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\frac{1}{1+\sqrt{a}}\right]^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(\frac{1}{a+2\sqrt{a}+1}\right)\)
\(=\frac{\left(a+2\sqrt{a}+1\right)}{a+2\sqrt{a}+1}\)
\(=1=VP\)
Vậy đẳng thức được chứng minh
Chứng minh các đẳng thức sau
a) \(\left(\frac{2\sqrt{6}-\sqrt{3}}{2\sqrt{2}-1}+\frac{5+2\sqrt{5}}{2+\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
b) \(\frac{a-b}{b^2}\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}=-a\)(Với b<a<0
c)\(\left(\sqrt{a}+\frac{1-a\sqrt{a}}{1-\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\)với a\(\ge0\),a khác 1
d) \(\left(\frac{3\sqrt{5}-\sqrt{15}}{\sqrt{27}-3}+\frac{2\sqrt{5}}{\sqrt{3}}\right)40\sqrt{15}=600\)
e) \(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)với x\(\ge0;x\ne1\)
chứng minh đẳng thức
\(\left(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{a-b}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\right).\frac{1}{\sqrt{a}+\sqrt{b}}=1\)
với a\(\ge0\)
ĐK: \(a,b\ge0,a\ne b\)
\(A=\left(\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\sqrt{a}+\sqrt{b}-\sqrt{ab}\right).\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(A=\left(\sqrt{ab}+\sqrt{a}+\sqrt{b}-\sqrt{ab}\right).\frac{1}{\sqrt{a}+\sqrt{b}}\)
\(A=\left(\sqrt{a}+\sqrt{b}\right).\frac{1}{\sqrt{a}+\sqrt{b}}=1=VP\)
Vậy đẳng thức được cm.
1 a)Chứng minh đẳng thức \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)với \(a\ge0\) và \(a\ne1\)
b) Tìm giátrị nhỏ nhất củabiểu thức P=\(\sqrt{x^2+6x+2011}\)
Bài 1:
a) \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}\cdot\frac{\sqrt{a}-1-a+\sqrt{a}}{\sqrt{a}-1}\)
\(=\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}\cdot\frac{-a+2\sqrt{a}-1}{\sqrt{a}-1}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}\cdot\frac{-\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\)
\(=-\left(a-1\right)\)
\(=1-a\)
b) \(P=\sqrt{x^2+6x+2011}\)
\(P=\sqrt{x^2+6x+9+2002}\)
\(P=\sqrt{\left(x+3\right)^2+2002}\ge\sqrt{2002}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\)
. Chứng minh đẳng thức
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}=\sqrt{2}-1\) b) \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}=1\)
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) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
chứng minh đẳng thức:
\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\)
Ta có :
\(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)\)
\(=4-a=VP\)
=> đpcm
Bổ sung ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)dùm mình nhé ;-;