\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\frac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\)
\(=a-\sqrt{ab}+b-\sqrt{ab}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2\)
chứng minh:
\(\frac{a\sqrt{a}+b
}{a-b}.\sqrt{\frac{ab+b^2-2\sqrt{ab^2}}{a.\left(a+2\sqrt{b}\right)+b}}.\left(\sqrt{a}+\sqrt{b}\right)=b\left(a>b>0\right)\)
chứng minh câu đẳng thức
1)\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
2)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
3)\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}=1\)(a lớn hơn bằng 0,b lớn hơn bằng 0)
4)\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)(a lớn hơn bằng 0,a khác 1)
help me:<<<
Chứng minh rằng nếu a,b>0 ta luôn có: \(\frac{a+2\sqrt{ab}+9b}{\sqrt{a}+3\sqrt{b}-2\sqrt[4]{ab}}-2\sqrt{b}=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
Chứng minh rằng ,nếu a,b>0 thi ta có:
\(\frac{a+2\sqrt{ab}+9b}{\sqrt{a}+3\sqrt{b}+2\sqrt[4]{ab}}-2\sqrt{b}=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
Chứng minh \(\frac{a\sqrt{b}+b}{a-b}\sqrt{\frac{ab+b^2-2\sqrt{ab^2}}{a\left(a+2\sqrt{b}+b\right)}}\left(\sqrt{a}+\sqrt{b}\right)=b\left(a>b>0\right)\)
Chứng minh rằng nếu a,b>0 thì ta luôn có
\(\frac{a+2\sqrt{ab}+9b}{\sqrt{a}+3\sqrt{b}-2\sqrt[4]{ab}}-2\sqrt{b}=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
Chứng minh \(\frac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\frac{ab+b^2-2\sqrt{ab^3}}{a\cdot\left(a+2\sqrt{b}\right)+b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=b\) với a > b > 0
Chứng minh đẳng thức sau với a,b > 0
\(\frac{a+2\sqrt{ab}+9b}{\sqrt{a}+3\sqrt{b}-2\sqrt[4]{ab}}-2\sqrt{b}=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
C/m biểu thức
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)=1\)(a,b>0,a\(\ne\)0
b)\(\frac{a-b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\left(a,b>0,a\ne b\right)\)
c)\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\left(a>0,a\ne1\right)\)
d)\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1-a\right)^2\left(a\ge0,a\ne1\right)\)
Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
