ta có (a-b)² >= 0 V  a,b

(=) a² -2ab+b² >=0

(=) a² + b² >= 2ab

(=) (a² + b²)/2 >= ab(ĐPCM)

#Học-tốt

\(\frac{a^2+b^2}{2}\ge ab\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) * đúng *

ban vao day nha nho   ti ck   cho mk nha!!!

https://lazi.vn/edu/exercise/cho-hinh-vuong-abcd-ve-goc-xay-90-do-ax-cat-bc-o-m-ay-cat-cd-o-n-a-cm-man-vuong-can

(=) (x-2)(x-2+x-3)=0

(=) (x-2)(2x-5)=0

(=) x-2=0 hoặc 2x-5=0

bn tự giải nha

#Học-tốt

\(\left(x-2\right)^2+\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2+x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\)

\(\Leftrightarrow x-2=0hoAC2x-5=0\)

\(\Leftrightarrow x\in\left\{2;\frac{5}{2}\right\}\)

Thay t = 3 vào phương trình, ta được:

\(1-a-3=2a\left(a+2\right)\)

\(\Leftrightarrow-2-a=2a^2+4a\)

\(\Leftrightarrow2a^2+5a+2=0\)

Ta có \(\Delta=5^2-4.2.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}a=\frac{-5+3}{4}=\frac{-1}{2}\\a=\frac{-5-3}{4}=-2\end{cases}}\)

\(2\left(x^4+y^4\right)\ge xy^3+x^3y+2x^2y^2\)

\(\Leftrightarrow\left(x^4-2x^2y^2+y^4\right)+\left(x^4-x^3y\right)+\left(y^4-xy^3\right)\ge0\)

\(\Leftrightarrow\left(x^2-y^2\right)^2+x^3\left(x-y\right)+y^3\left(y-x\right)\ge0\)

\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\) 

\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{2}\right]\ge0\) ( đúng )

\(\left(4x+3\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow\left(4x+3\right)^2=\left(2x-2\right)^2\)

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow12x^2+2x+30x+5=0\)

\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)

\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}6x+1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy tập hợp nghiệm của pt \(S=\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

\(\left(4x+3\right)^2=4.\left(x-1\right)^2\)

\(\Leftrightarrow16x^2+24x+9=4.\left(x^2-2x+1\right)\)

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow12x^2+30x+2x+5=0\)

\(\Leftrightarrow6x\left(2x+5\right)+\left(2x+5\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\6x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-1}{6}\end{cases}}\)

Thay x = 4 vào phương trình, ta được :

\(1-m=2\left(2m+1\right)\left(m-1\right)\)

\(\Leftrightarrow2\left(2m+1\right)\left(m-1\right)+\left(m-1\right)=0\)

\(\Leftrightarrow\left(m-1\right)\left(4m+2+1\right)=0\)

\(\Leftrightarrow\left(m-1\right)\left(4m+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}m-1=0\\4m+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=1\\m=\frac{-3}{4}\end{cases}}\)

Hằng đẳng thức:\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Khi đó:

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

\(=a+b+c\)

\(=2011\)