Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
.
.
.
.
.
.
...
.
.
..
.
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
..
.
..
.
..
.
.
.
.
.
.
.
.
.
Hello
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
\(a,B=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}\right):\left(\frac{1-xy+x+y+2xy}{1-xy}\right)\)
\(B=\frac{\sqrt{x}+\sqrt{y}+x\sqrt{y}+y\sqrt{x}+\sqrt{x}-\sqrt{y}-x\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{1+xy+x+y}\)
\(B=\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(y+1\right)+\left(y+1\right)}\)
\(B=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}\)
\(B=\frac{2\sqrt{x}}{x+1}\)
\(b,B=\frac{2\sqrt{\frac{2}{2+\sqrt{3}}}}{\frac{2}{2+\sqrt{3}}+1}\)
\(\frac{2\sqrt{\frac{4}{4+2\sqrt{3}}}}{\frac{4}{4+2\sqrt{3}}+1}\)
\(B=\frac{2\sqrt{\frac{4}{\left(\sqrt{3}+1\right)^2}}}{\frac{4}{\left(\sqrt{3}+1\right)^2}+1}\)
\(B=\frac{2.2}{\sqrt{3}+1}:\frac{4+2\sqrt{3}}{\sqrt{3}+1}\)
\(B=\frac{4}{\left(\sqrt{3}+1\right)^2}\)
\(B=\left(\frac{2}{\sqrt{3}+1}\right)^2\)
\(c,B=\frac{2\sqrt{x}}{x+1}\)
\(B=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}}\)
ta có :
\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)
dấu "=" xảy ra khi \(x=1\)
\(< =>MAX:B=\frac{2}{2}=1\)
Đk: x \(\ge\)0; y \(\ge\)0; xy \(\ne\)1
Ta có: B = \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\frac{x+y+2xy}{1-xy}\right)\)
B = \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\frac{1-xy+x+y+2xy}{1-xy}\)
B = \(\frac{x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}\cdot\frac{1-xy}{x+y+xy+1}\)
B = \(\frac{2\sqrt{x}+2y\sqrt{x}}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{2\sqrt{x}}{x+1}\)
b) Ta có: \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)
=> \(x=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)=> \(\sqrt{x}=\sqrt{3}-1\)
Do đó, B = \(\frac{2.\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\frac{2\sqrt{3}-2}{5-2\sqrt{3}}=\frac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\frac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)
B = \(\frac{6\sqrt{3}+2}{13}\)
c) Ta có: \(\frac{1}{B}=\frac{x+1}{2\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\ge2\cdot\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{2\sqrt{x}}}=2\cdot\sqrt{\frac{1}{4}}=1\)(đk: x \(\ne\)0)
=> \(B\le\frac{1}{1}=1\)Dấu "==" xảy ra<=> \(\frac{\sqrt{x}}{2}=\frac{1}{2\sqrt{x}}\) => \(2\sqrt{x}=2\) => \(x=1\)
3)\(...=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\right].\frac{1-xy}{x+xy}\)
= \(\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{x\left(1+y\right)}\)= \(\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(1+y\right)}=\frac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}=\frac{2}{\sqrt{x}}\)
