a) Đúng

b) \(\sqrt{1^3+2^3+3^3+4^3}=1+2+3+4\)

\(\sqrt{1^3+2^3+3^3+4^3+5^3=1+2+3+4+5}\)

Các đẳng thức trên luôn đúng:

Ta có công thức tổng quát

\(\sqrt{1^3+2^3+...+n^3}=1+2+..+n\)

VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)

\(=10-1=9=VP\)

Cảm ơn bạn nhé dù mình biết đáp án rồi :)

a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

\(\frac{4.\left(\sqrt{3}+1\right)}{\sqrt{3}-1}-\frac{2+\sqrt{3}}{2-\sqrt{3}}\)

\(\Leftrightarrow\frac{4\left(\sqrt{3}+1\right)\left(2-\sqrt{3}\right)}{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}-\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(2-\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)

\(\Leftrightarrow\frac{4\left(\sqrt{3}+1\right)\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}\)

\(\Rightarrow\frac{3\sqrt{3}-5}{3\sqrt{5}-5}=1\left(đpcm\right)\)

a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)

--> dpcm

b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)

=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

=\(-1.\left(7-5\right)\)

=-1.2

=-2

1: \(\frac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}\)

\(=\frac{\sqrt2\left(2+\sqrt3\right)}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2\left(2-\sqrt3\right)}{2-\sqrt{4-2\sqrt3}}\)

\(=\frac{\sqrt2\left(2+\sqrt3\right)}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2\left(2-\sqrt3\right)}{2-\sqrt{\left(\sqrt3-1\right)^2}}\)

\(=\frac{\sqrt2\left(2+\sqrt3\right)}{2+\sqrt3+1}+\frac{\sqrt2\left(2-\sqrt3\right)}{2-\sqrt3+1}=\frac{\sqrt2\left(2+\sqrt3\right)}{3+\sqrt3}+\frac{\sqrt2\left(2-\sqrt3\right)}{3-\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\frac{2\left(2+\sqrt3\right)}{\sqrt3\left(\sqrt3+1\right)}+\frac{2\left(2-\sqrt3\right)}{\sqrt3\left(\sqrt3-1\right)}\right)=\frac{1}{\sqrt2}\cdot\left(\frac{4+2\sqrt3}{\sqrt3\left(\sqrt3+1\right)}+\frac{4-2\sqrt3}{\sqrt3\left(\sqrt3-1\right)}\right)\)

\(=\frac{1}{\sqrt2}\left\lbrack\frac{\left(\sqrt3+1\right)^2}{\sqrt3\left(\sqrt3+1\right)}+\frac{\left(\sqrt3-1\right)^2}{\sqrt3\left(\sqrt3-1\right)}\right\rbrack=\frac{1}{\sqrt2}\cdot\frac{\sqrt3+1+\sqrt3-1}{\sqrt3}=\frac{1}{\sqrt2}\cdot2=\sqrt2\)

2: \(\left(\sqrt{x}-\frac{x}{x+\sqrt{x}}\right):\frac{\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)

\(=\left(\sqrt{x}-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{\sqrt{x}\left(x-1\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\cdot\sqrt{x}\left(\sqrt{x}+1\right)=x\sqrt{x}\)