Bài 2:

\(\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

Ta có: \(\tan\alpha.\cot\alpha=1\Rightarrow\tan\alpha=\frac{1}{\cot\alpha}\)

Đặt \(\cot\alpha=t\)thì \(\tan\alpha=\frac{1}{t}\)

Khi đó \(B=\frac{1}{1+\frac{1}{t}}+\frac{1}{1+t}=\frac{t}{t+1}+\frac{1}{1+t}=1\)

1+tan a=1+sina/cosa = sina+cosa/cosa

1+cota=sina+cosa/sina

=>B=1.

bài này giải thế này nhé

Áp dụng bất đẳng thức 

\(a^3+b^3\ge ab\left(a+b\right)\)\(a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\)

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

\(=\frac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\frac{x+1}{x+\sqrt{x}+1}=\frac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}.\frac{1}{x+\sqrt{x}+1}=\frac{-\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)

Ta có \(x^2-4\left(m-1\right)x+5=0\)    \(\left(a=1;b=-4\left(m-1\right);c=5\right)\)

a) Vì pt có nghiệm x=1\(\Rightarrow a+b+c=0\)

                                     \(\Leftrightarrow1-4\left(m-1\right)+5=0\)

                                     \(\Leftrightarrow1-4m+4+5=0\)

                                      \(\Leftrightarrow4m=10\)

                                      \(\Leftrightarrow m=\frac{5}{2}\)

b) Vì pt có nghiệm x1=1\(\Rightarrow x2=\frac{c}{a}=5\)

ĐK: a,b>0 , a khác b

\(A=\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}.\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}}\right]:\left(\frac{a^2-b^2}{ab}\right)\)

\(=\frac{a-b}{b}:\frac{\left(a-b\right)\left(a+b\right)}{ab}=\frac{a-b}{b}.\frac{ab}{\left(a-b\right)\left(a+b\right)}=\frac{a}{a+b}\)

Với b=1, A=2 ta có: 

\(\frac{a}{a+1}=2\Leftrightarrow a=2a+2\Leftrightarrow a=-2\) loại 

vậy không tồn tại a để A=2 b=1

\(A=\left[\left(\sqrt{\frac{a}{b}}-1\right).\left(\sqrt{\frac{a}{b}}+1\right)\right]:\left(\frac{a}{b}-\frac{b}{a}\right)\)

\(A=\left[\left(\sqrt{\frac{a}{b}}\right)^2-1\right]:\left(\frac{a^2}{ab}-\frac{b^2}{ab}\right)\)

\(A=\left(\frac{a}{b}-1\right):\left[\frac{\left(a-b\right)\left(a+b\right)}{ab}\right]\)

\(A=\left(\frac{a-b}{b}\right).\left[\frac{ab}{\left(a-b\right)\left(a+b\right)}\right]\)

\(A=\frac{a}{a+b}\)