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a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
\(=\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}.\frac{\sqrt{a}+1}{\sqrt{b}-1}}=\sqrt{\frac{a-1}{b-1}}=\sqrt{\frac{7,25-1}{3,25-1}}=\sqrt{\frac{625}{225}}=\frac{5}{3}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}-1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}=\frac{\sqrt{\sqrt{a}-1}}{\sqrt{\sqrt{b+1}}}\cdot\frac{\sqrt{\sqrt{a}+1}}{\sqrt{\sqrt{b}-1}}\)
\(=\frac{\sqrt{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}}{\sqrt{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}}=\frac{\sqrt{a-1}}{\sqrt{b-1}}\)
Thay a = 7,25 ; b = 3,25 ta có
\(\frac{\sqrt{7,25-1}}{\sqrt{3,25-1}}=\frac{\sqrt{6,25}}{\sqrt{2,25}}=\frac{2,5}{1,5}\)
Đúng nha
a/ Với x = \(23-12\sqrt{3}\) ta có:
\(x-11=23-12\sqrt{3}-11=12-12\sqrt{3}=12\left(1-\sqrt{3}\right)\)
\(\sqrt{x-2}-3=\sqrt{23-12\sqrt{3}-2}-3=\sqrt{21-12\sqrt{3}}-3=\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-3=\sqrt{\left(3-2\sqrt{3}\right)^2}-3=2\sqrt{3}-6\) \(=2\sqrt{3}\left(1-\sqrt{3}\right)\)
=>\(\frac{x-11}{\sqrt{x-2}-3}=\frac{12\left(1-\sqrt{3}\right)}{2\sqrt{3}\left(1-\sqrt{3}\right)}=\frac{12}{2\sqrt{3}}=\frac{2\sqrt{3}.2\sqrt{3}}{2\sqrt{3}}=2\sqrt{3}\)
b/ \(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}=\frac{1-\sqrt{a}}{2\left(1-a\right)}+\frac{1+\sqrt{a}}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}\)
=\(\frac{2}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{1-a+a^2-a^2-2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{-a-1}{1-a^3}\)
Thay : \(a=\sqrt{2}tacó:\)
\(\frac{-\sqrt{2}-1}{1-\sqrt{2}^3}=\frac{-\left(1+\sqrt{2}\right)}{1-2\sqrt{2}}\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a. ĐK:\(a,b\ge0;b\ne1\)
\(=\sqrt{\frac{a-1}{b-1}}\)
Thay a=7,25;b=3,25:
\(=\sqrt{\frac{6,25}{2,25}}=\frac{5}{3}\)
b. ĐK:\(x\ge-2\)
\(=4x-2\sqrt{2}+\sqrt{x^2\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+x\sqrt[4]{x+2}\)
Thay x=-\(\sqrt{2}\) vào rồi tính.
Thanks Nguyen nhiều nhiều nhiều <3