170

280

210

Đây là một dẫn chứng cho sự giàu đẹp của Tiếng Việt

\(2x^2-6x\\ =2x^2-6x+\dfrac{9}{2}-\dfrac{9}{2}\\ =2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ =2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{2}\\ =2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

\(\text{Ta có: }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)

Dấu \("="\) xảy ra khi:

\(2\left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow \left(x-\dfrac{3}{2}\right)^2=0\\\Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(GTNN\) của biểu thức là \(-\dfrac{9}{2}\) khi \(x=\dfrac{3}{2}\)

(3x-1)+(1-3x)=6

Suy ra: 3x-1+1-3x=6

3x-3x=6

9x=6

x=6:9

x=2/3

\(\text{b) }\left(x+3\right)^2-\left(x+1\right)\left(x-1\right)\\ =\left(x+3\right)^2-\left(x^2-1^2\right)\\ =x^2+2\cdot x\cdot3+3^2-x^2+1\\ =\left(x^2-x^2\right)+6x+\left(9+1\right)\\ =6x+10\\ \)

\(\text{c) }\left(x-5\right)^2-\left(x+2\right)^2\\ =\left(x^2-2\cdot x\cdot5+5^2\right)-\left(x^2+2\cdot x\cdot2+2^2\right)\\ =x^2-10x+25-x^2-4x-4\\ =\left(x^2-x^2\right)-\left(10x+4x\right)+\left(25-4\right)\\ =-14x+21\\ \)

\(\text{d) }\left(x+3\right)^2-\left(x-3\right)^2\\ =\left(x^2+2\cdot x\cdot3+3^2\right)-\left(x^2-2\cdot x\cdot3+3^2\right)\\ =x^2+6x+9-x^2+6x-9\\ =\left(x^2-x^2\right)+\left(6x+6x\right)+\left(9-9\right)\\ =12x\\ \)

\(\text{e) }2x\left(x+1\right)-\left(x+3\right)^2-x^2\\ =2x^2+2x-\left(x^2+2\cdot x\cdot3+3^2\right)-x^2\\ =2x^2+2x-x^2-6x-9-x^2\\ =\left(2x^2-x^2-x^2\right)+\left(2x-6x\right)-9\\ =-4x-9\\ \)

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

Bạn trình bày cho rõ ràng xem nào.

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

Thay \(a+b+c=0\) vào biểu thức ta được:

\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\\ =0\left(a^2+b^2+c^2-ab-ac-bc\right)\\ =0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a^3+b^3+c^3\right)-3abc=0\\a^3+b^3+c^3=3abc\left(đpcm\right) \)

Vậy.....................

\(S=\left|x+3\right|+\left|x-17\right|\\ S=\left|x+3\right|+\left|17-x\right|\)

Áp dụng bất đẳng thức : \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta được:

\(S=\left|x+3\right|+\left|17-x\right|\ge\left|\left(x+3\right)+\left(17-x\right)\right|\\ S\ge\left|x+3+17-x\right|\\ S\ge\left|20\right|\\ S\ge20\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}x+3\ge0\\x-17\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le17\end{matrix}\right.\Leftrightarrow-3< x< 17\)

Vậy \(S_{\left(Min\right)}=20\) khi \(-3< x< 17\)