\(P=x+\frac{1}{x-2}\left(x>2\right)\)

\(=x-2+\frac{1}{x-2}+2\)

Vì x > 2 => x - 2 > 0

                  1 / x-2 > 0

Áp dụng BĐT cô - si cho hai số dương:

\(x-2+\frac{1}{x-2}\ge2\sqrt{\left(x-2\right).\frac{1}{x-2}}=2\)

\(\Rightarrow x-2+\frac{1}{x-2}\ge2+2=4\)

\(\Rightarrow P\ge4\Rightarrow MinP=4\Leftrightarrow x-2=\frac{1}{x-2}\Leftrightarrow\left(x-2\right)^2=1\Leftrightarrow x=3\)

\(P=x-2+\frac{1}{x-2}+2\ge2+2=4\)

Đẳng thức xảy ra khi x - 2 = 1 hay x = 3

P/s: Đúng ko ak?

Làm sao CM 1 điều hiển nhiên được, tan 15 độ = \(2-\sqrt{3}\)thì ai cũng phải công nhận

Ta có:

\(VT=\tan15^o=2-\sqrt{3}=VP\left(đpcm\right)\)

\(4x^2-8x+3\)

\(=4x^2-6x-2x+3\)

\(=2x\left(2x-3\right)-\left(2x-3\right)\)

\(=\left(2x-1\right)\left(2x-3\right)\)

\(4x^2-8x+3\)

\(=4x^2-6x-2x+3\)

=\(2x\left(2x-3\right)-\left(2x-3\right)\)

\(=\left(2x-1\right)\left(2x-3\right)\)

Ta có ; 

        y =\(2+\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)

Mà \(\left(\sqrt{2}x-\sqrt{2}\right)^{2^{ }^{ }}+3\ge3\)

\(\Rightarrow\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge\sqrt{3}\)

\(\Rightarrow2+\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge2+\sqrt{3}\)

\(\Rightarrow y\ge2+\sqrt{3}\)

Vậy gía trị nhỏ nhất của biểu thức là 2+\(\sqrt{3}\).Dấu "=" xảy ra khi x=1

Nếu đề bài là

Tính P=\(\frac{x_1^2+x_1-1}{x_1}\)-\(\frac{x_2^2+x_2-1}{x_2}\)

Thì lời giải như sau:

Theo định lý Viete, ta có:

x₁.x₂=-1

Khi đó P=\(\frac{x_1^2+x_1+x_1.x_2}{x_1}\)-\(\frac{x_2^2+x_2+x_1.x_2}{x_2}\)

Do x₁ và x₂ không thể bằng không nên ta chia tử mẫu của mỗi hạng tử cho x₁,x₂

Khi đó P=x₁+x₂+1-(x₂+x₁+1)=0

\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)

\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)

Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)

=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)

\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)

Ta có: 21=\(\sqrt{441}\)

Mà \(\sqrt{441}>\sqrt{147}\)

\(\Rightarrow\)21>\(\sqrt{147}\)

Ta có:

\(21=\sqrt{441}\)

Vì 441 > 147

\(\Rightarrow\sqrt{441}>\sqrt{147}\)

\(\Rightarrow21>\sqrt{147}\)

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

\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)

b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)

\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)

c) (ko bt có sai đề ko, làm mãi ko ra) 

d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)

\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)

\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)

\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)

\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)