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

\(=x^2-6x+9+\left(5x-x^2+10-2x\right)\)

\(=x^2-6x+9+3x-x^2+10\)

\(=-3x+19\)

Ta co:

\(A=\left(x+z\right)\left(y+t\right)\le\frac{\left(x+y+z+t\right)^2}{4}\le\frac{4\left(x^2+y^2+z^2+t^2\right)}{4}=1\)

Dau '=' xay ra khi \(x=y=z=t=\frac{1}{2}\)

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

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

+) Đặt \(B=\left|x-1\right|+\left|4-x\right|\ge\left|x-1+4-x\right|=3\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left(x-1\right)\left(4-x\right)=0\)

\(\Leftrightarrow1\le x\le4\)

+) Đặt \(C=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)

Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow2\le x\le3\)

\(\Rightarrow A=\left|x-1\right|+\left|4-x\right|+\left|x-2\right|+\left|3-x\right|\ge4\)

Dấu '' = '' xảy ra

\(\Leftrightarrow\hept{\begin{cases}1\le x\le4\\2\le x\le3\end{cases}\Leftrightarrow2\le x\le3}\)

Vậy.................

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Hoặc bạn vào trong câu hỏi tương tự nha !

9x² - 4 = (2x - 1)(3x + 2)

=> (3x - 2)(3x + 2) - (2x - 1)(3x + 2) = 0

=> (3x + 2)(3x - 2 - 2x + 1) = 0

=> (3x + 2)(x - 1) = 0

=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

\(x^2-4x+4-x^2-3x-3x+15=1\)

\(-10x+19=1\)

\(-10x=-18\)

\(x=\frac{9}{5}\)

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

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

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

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

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

xong!

cảm ơn nha

\(x^2-y^2+6x+9\)

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

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

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

x^2-5x+2xy-10y
phân tích đa thức thành nhân tử

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

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

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

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