(2x-3)^2-(x+5)^2=2x-3=x+5=>x=8

abcd là HÌNH THANG THƯỜNG nha

có chép nhầm k bn

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

Xét vế phải:

\(VP=-x^2-2x+2\)

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

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

Ta có: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+1\right)^2+3\le3\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x=-1\)(1)

Xét vế trái:

\(VT=\left|x-1\right|+\left|x+2\right|\)

\(VT=\left|1-x\right|+\left|x+2\right|\)

Ta có: \(\left|1-x\right|+\left|x+2\right|\ge\left|1-x+x+2\right|=\left|3\right|=3\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}1\ge x\\x\ge-2\end{cases}}}\)(2)

Từ (1) ;(2) ta có: 

\(VT=\left|x-1\right|+\left|x+2\right|\ge3\ge VP\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}VT=3\\VP=3\end{cases}\Leftrightarrow\hept{\begin{cases}-2\le x\le1\\x=-1\end{cases}}\Leftrightarrow x=-1}\)

Vậy \(x=-1\)

Tham khảo nhé~

a, k ph đc

b,Đặt \(A=...=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt x^2+5x+4=t,ta có:

\(A=t\left(t+2\right)-24=t^2+2t-24=t^2-4t+6t-24=t\left(t-4\right)+6\left(t-4\right)=\left(t-4\right)\left(t+6\right)\)

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

a)  \(9x^2+6x-8\)

\(=9x^2+12x-6x-8\)

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

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

b) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

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

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

c) \(x^8+x^7+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

a, \(A=x^2-6x+11\)

\(=x^2-2.3.x+9+2\)

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

Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(MinA=3\Leftrightarrow x=3\)

b, \(B=2x^2+10x-1\)

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

\(=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{21}{4}\)

\(=2\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\)

Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MinB=-\frac{21}{4}\Leftrightarrow x=-\frac{5}{2}\)

c, \(C=5x-x^2\)

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

\(=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{25}{4}\)

\(=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\)

Ta có: \(-\left(x+\frac{5}{2}\right)^2\le0\Leftrightarrow-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MaxB=\frac{25}{4}\Leftrightarrow x=-\frac{5}{2}\)