\(B=-2x^2+4x-5\)

\(=-2\left(x^2-2x+\frac{5}{2}\right)\)

\(=-2\left(x^2-2x+1+\frac{3}{2}\right)\)

\(=-2\left[\left(x-1\right)^2+\frac{3}{2}\right]\)

\(=-2\left[\left(x-1\right)^2\right]-3\le3< 0\forall x\)

\(B=-2x^2+4x-5\)

\(B=-2\left(x^2-2x+\frac{5}{2}\right)\)

\(B=-2\left(x^2-2x+1+\frac{3}{2}\right)\)

\(B=-2\left[\left(x-1\right)^2+\frac{3}{2}\right]\)

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

Mà \(\hept{\begin{cases}-2\left(x-1\right)^2\le0\forall x\\-3< 0\end{cases}\Rightarrow B< 0\forall x}\)

a) \(2x^2+4x+2\)

\(=2\left(x^2+2x+1\right)\)

\(=2\left(x^2+x+x+1\right)\)

\(=2\left[x\left(x+1\right)+\left(x+1\right)\right]\)

\(=2\left(x+1\right)^2\)

b) \(9\left(2x-5\right)-6y\left(5-2x\right)\)

\(=9\left(2x-5\right)+6y\left(2x-5\right)\)

\(=\left(9+6y\right)\left(2x-5\right)\)

\(=3\left(3+2y\right)\left(2x-5\right)\)

\(A=\left(2x-1\right)^2-\left(5+x\right)\left(5-x\right)+4x\)

\(=4x^2-4x+1-\left(25-x^2\right)+4x\)

\(=4x^2-4x+1-25+x^2+4x\)

\(=5x^2-24\)

Thay x = -2 vào bt ,ta được: \(5.\left(-2\right)^2-24=-4\)

\(3x^2\left(x^2-5x+2\right)\)

\(=3x^4-15x^3+6x^2\)

là 3:5:9 nha, mình ghi nhầm

sinh hỏ??? m đăng lên cái diễn đàn khác đi

Nhìn qua thấy bậc của bđt là không đồng bậc nên hơi căng đấy...

Chú ý: \(2019=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{x+y+z}{xyz}\Rightarrow xyz=\frac{x+y+z}{2019}\)

\(LHS=\Sigma_{cyc}\frac{\sqrt{2019x^2+1}+1}{x}=\Sigma_{cyc}\frac{\sqrt{\frac{x}{y}+\frac{x^2}{yz}+\frac{x}{z}+1}+1}{x}\)( thay \(2019=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\))

\(=\Sigma_{cyc}\frac{\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}+1}{x}=\Sigma_{cyc}\left[\sqrt{\frac{\left(\frac{x}{y}+1\right)}{x}.\frac{\left(\frac{x}{z}+1\right)}{x}}+\frac{1}{x}\right]\)

\(=\Sigma_{cyc}\sqrt{\left(\frac{1}{y}+\frac{1}{x}\right)\left(\frac{1}{z}+\frac{1}{x}\right)}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{2}\left[4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(=3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{3\left(xy+yz+zx\right)}{\frac{\left(x+y+z\right)}{2019}}=\frac{6057\left(xy+yz+zx\right)}{x+y+z}\)

\(\le\frac{6057.\frac{\left(x+y+z\right)^2}{3}}{x+y+z}=2019\left(x+y+z\right)\)(đpcm)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{\frac{3}{2019}}\)

P/s: Check hộ t phát:3

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì bài toán thành

Cho: \(ab+bc+ca=2019\)

Chứng minh:

\(\sqrt{2019+a^2}+\sqrt{2019+b^2}+\sqrt{2019+c^2}+\left(a+b+c\right)\le2019\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Ta có:

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

\(VT=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}+\left(a+b+c\right)\)

\(\le3\left(a+b+c\right)\)

\(VP=\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

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

\(\ge3\left(a+b+c\right)\)

Tới đây bí :(

\(x^2-\left(x-2\right)\left(2x+1\right)=4\)

\(\Leftrightarrow\left(x-2\right)\left(2x+1\right)=\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow2x+1=x+2\)

\(\Leftrightarrow2x+1-x-2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Dòng thứ 2 của em nếu x - 2 =0 thì sao?

Bài giải:

pt <=> \(x^2-4-\left(x-2\right)\left(2x+1\right)=0\)

<=> \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(2x+1\right)=0\)

<=> \(\left(x-2\right)\left(x+2-2x-1\right)=0\)

<=> \(\left(x-2\right)\left(1-x\right)=0\)

<=> x -2 =0 hoặc  1- x =0

<=> x = 2 hoặc x = 1. 

Vậy ...