khó ha

\(bđt< =>\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(< =>a^2+2ab+b^2\ge4ab\)

\(< =>a^2+b^2\ge2ab\)

\(< =>\left(a-b\right)^2\ge0\)*đúng*

Vậy ta có điều phải chứng minh

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}\ge0\)

\(\frac{b}{ab}+\frac{a}{ab}-\frac{4}{a+b}\ge0\)

\(\frac{a+b}{ab}-\frac{4}{a+b}\ge0\)

\(\frac{\left(a+b\right)^2}{ab\left(a+b\right)}-\frac{4ab}{ab\left(a+b\right)}\ge0\)

\(a^2+2ab+b^2-4ab\ge0\Leftrightarrow\left(a-b\right)^2\ge0\)

Đăngr thức xảy ra <=> a = b 

,[x-y+z]^2+[z-y]^2+2.[x-y+z][y-z] (x - y + z)² + (z - y)² + 2(x - y + z)(y - z)
= (x - y + z)² + 2(x - y + z)(y - z) + (y - z)²
= (x - y + z + y - z)²
= x²

Ta có:

\(\left(x-y+z\right)^2+\left(z-y\right)^2+2.\left(x-y+z\right).\left(y-z\right)\)

\(=\left(x-y+z\right)^2+2.\left(x-y+z\right).\left(y-z\right)+\left(y-z\right)^2\)

\(=\left(x-y+z+y-z\right)^2\)

\(=x^2\)

Học tốt nhé 

Ta có: \(y'=a\)\(cosx-b\)\(sinx+1\)

y đồng biến trên R \(\Leftrightarrow y'\ge0,\forall x\in R\)

                         \(\Leftrightarrow acosx-bsinx+1\ge0,\forall x\in R\)(*)

Theo bất đẳng thức Schwartz thì:

\(|acosx-bsinx|\le\sqrt{a^2+b^2},\forall x\)

\(\Leftrightarrow-\sqrt{a^2+b^2}\le acos-bsinx\le\sqrt{a^2+b^2},\forall x\)

\(\Leftrightarrow1-\sqrt{a^2+b^2}\le acos-bsinx+1\le1+\sqrt{a^2+b^2},\forall x\)

Do đó (*) \(\Leftrightarrow1-\sqrt{a^2+b^2}\ge0\)

\(\Leftrightarrow\sqrt{a^2+b^2}\le1\)

\(\Leftrightarrow a^2+b^2\le1\)

Ta có : \(h_1=13,6h_2\)=> \(\frac{h_1}{h_2}=\frac{13,6}{1}\)hay \(\frac{h_1}{13,6}=\frac{h_2}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{h_1}{13,6}=\frac{h_2}{1}=\frac{h_1+h_2}{13,6+1}=\frac{0,44}{14,6}=\frac{11}{365}\)

=> \(h_1=\frac{11}{365}\cdot13,6=\frac{748}{1825}\)

\(h_2=\frac{11}{365}\cdot1=\frac{11}{365}\)

P/S : Đề như thế nào?Số dữ quá 

Ta có :

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=\left(-a\right)^2\)

\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

Tương tự  \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

Mặt khác \(\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^3+3bc\left(b+c\right)+c^3=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow P=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ba}=\frac{a^3+b^3+c^3}{2abc}\)

\(=\frac{3abc}{2abc}=\frac{3}{2}\)

Vậy P = 3/2 

P=3/2

Ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)(Vì a+b+c=0)

Khi đó ta có:\(N=\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ba}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)

Vậy N=3

\(N=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}\)

\(< =>N-1=\frac{a^3+b^3+c^3-abc}{abc}=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}\)

Do \(a+b+c=0\)\(< =>N-1=\frac{0.\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}=0\)

\(< =>N=1\)

Đặt \(f\left(x\right)=2arctgx+arcsin\left(\frac{2x}{1+x^2}\right)=\eta,x\ge1\)

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

\(=\frac{2}{1+x^2}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)^2}.\sqrt{\frac{\left(1+x^2\right)^2}{\left(1+x^2\right)^2-4x^2}}\)

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

\(=\frac{2}{1+x^2}+\frac{2\left(1-x^2\right)}{1+x^2}.\frac{1}{x^2-1}\left(x>1\Leftrightarrow x^2-1>0\right)\)

\(=\frac{2}{1+x^2}-\frac{2}{1+x^2}=0,\forall x>1\)

Suy ra f(x) là hằng số trên \(\left(1,\infty\right).\)

mà \(f\left(\sqrt{3}\right)=2arctg\sqrt{3}+arcsin\frac{\sqrt{3}}{2}=2.\frac{\eta}{3}+\frac{\eta}{3}=\eta\)

nên \(f\left(x\right)=\eta,\forall x\in\left(1,\infty\right)\)

Hơn nữa \(f\left(1\right)=2arctg1+arcsin1=2.\frac{\eta}{4}.\frac{\eta}{2}=\eta\)

Do vậy: \(2arctgx+arcsin\left(\frac{2x}{1+x^2}\right)=\eta,\forall x\ge1.\)

Ta có: a³ + b³ + c³ = 3abc

<=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

<=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

<=> (a + b + c)[(a + b)² - (a + b)c + c²) - 3ab(a + b + c) = 0

<=> (a + b + c)(a²  + 2ab + b² - ac - bc + c² - 3ab) = 0

<=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\left(loại\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{cases}}\)

<=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

<= > (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)<=> a = b = c

Khi đó: B = \(\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

ta có a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

<=> (a+b)³-3ab(a+b)+c³-3abc=0

<=> (a+b+c)³-3(a+b)c(a+b+c)-3ab(a+b+c)=0

<=> (a+b+c)(a²+b²+c²-ab-bc-ca)=0

<=> a²+b²+c²-ab-bc-ca=0 (vì a+b+c=0)

<=> (a-b)²+(b-c)²+(c-a)²=0

<=> a=b=c

khi đó \(B=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)

x2 - 5x - 2xy + 5y + y2 + 4

= (x2 - 2xy + y2) - (5x - 5y) + 4

= (x2 - xy - xy + y2) - 5.(x - y) + 4

= (x - y)2 - 5.1 + 4

= 1 - 5 + 4

= 0