c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

a) \(A=2x^2+2x+3\)

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

\(A=2\left[x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{5}{4}\right]\)

\(A=2\left[\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\right]\)

\(A=2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)

b) Biến đổi mẫu thức :

\(3x^2+4x+15\)

\(=3\left(x^2+\frac{4}{3}x+5\right)\)

\(=3\left[x^2+2\cdot x\cdot\frac{2}{3}+\left(\frac{2}{3}\right)^2+\frac{41}{9}\right]\)

\(=3\left[\left(x+\frac{2}{3}\right)^2+\frac{41}{9}\right]\)

\(=3\left(x+\frac{2}{3}\right)^2+\frac{41}{3}\)

\(B=\frac{5}{3\left(x+\frac{2}{3}\right)^2+\frac{41}{3}}\ge\frac{5}{\frac{41}{3}}=\frac{15}{41}\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{2}{3}=0\Leftrightarrow x=\frac{-2}{3}\)

c) \(C=-x^2+2x-2\)

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

\(C=-\left(x^2-2\cdot x\cdot1+1^2+1\right)\)

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

\(C=-1-\left(x-1\right)^2\le-1\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

d) Biến đổi mẫu thức tương tự câu b)

\(P=\frac{xy}{\left|xy\right|}+\frac{x-y}{\left|x-y\right|}\cdot\left(\frac{x}{\left|x\right|}-\frac{y}{\left|y\right|}\right)\)

TH1: \(x,y>0\)

+) Xét \(x>y\): \(P=\frac{xy}{xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{x}-\frac{y}{y}\right)=1+1\cdot\left(1-1\right)=1\)

+) Xét \(x< y\): \(P=\frac{xy}{xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{x}-\frac{y}{y}\right)=1+\left(-1\right)\cdot\left(1-1\right)=1\)

TH2: \(x,y< 0\)

+) Xét \(x>y\): \(P=\frac{xy}{xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{-x}-\frac{y}{-y}\right)=1+1\cdot\left[-1-\left(-1\right)\right]=1\)

+) Xét \(x< y\): \(P=\frac{xy}{xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{-x}-\frac{y}{-y}\right)=1\)

TH3: \(x>0;y< 0\): \(P=\frac{xy}{-xy}+\frac{x-y}{x-y}\cdot\left(\frac{x}{x}-\frac{y}{-y}\right)=-1+1\cdot\left(1+1\right)=1\)

TH4: \(x< 0;y>0\): \(P=\frac{xy}{-xy}+\frac{x-y}{y-x}\cdot\left(\frac{x}{-x}-\frac{y}{y}\right)=-1+\left(-1\right)\cdot\left(-1-1\right)=1\)

Nói chung với mọi x, y thì P = 1

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

+) \(x\ge-1\): \(A=\frac{x+1+2x}{\left(x-1\right)\left(3x+1\right)}=\frac{3x+1}{\left(x-1\right)\left(3x+1\right)}=\frac{1}{x-1}\)

+) \(x< -1\): \(A=\frac{-x-1+2x}{\left(x-1\right)\left(3x+1\right)}=\frac{x-1}{\left(x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)

Rồi thay x vào thôi :)) 

b/ \(3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow-101x=-303\)

\(\Rightarrow x=3\)

c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-79x=-158\)

\(\Rightarrow x=2\)

d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow-6x=5\)

\(\Rightarrow x=-\frac{5}{6}\)

e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow13x=130\)

\(\Rightarrow x=10\)

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

\(\Rightarrow A_{min}=-3\) khi \(x=2\)

\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)

\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)

\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\)

\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)

\(\Rightarrow C_{max}=21\) khi \(x=-4\)

\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)

\(\Rightarrow E_{max}=5\) khi \(x=2\)

Giups mik vs

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

Dấu \("="\Leftrightarrow x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

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

Dấu \("="\Leftrightarrow x=1\)

\(D=\dfrac{1}{-\left(x^2+2x+1\right)+6}=\dfrac{1}{-\left(x+1\right)^2+6}\ge\dfrac{1}{6}\)

Dấu \("="\Leftrightarrow x=-1\)

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

\(A_{min}=-7\) khi \(x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{3}{2}\)

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

\(C_{min}=-4\) khi \(x=1\)

Biểu thức D không tồn tại cả max lẫn min

1.

$A=2x^2-8x+1=2(x^2-4x+4)-7=2(x-2)^2-7$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A\geq 2.0-7=-7$

Vậy $A_{\min}=-7$ khi $x-2=0\Leftrightarrow x=2$

2.

$B=x^2+3x+2=(x^2+3x+1,5^2)-0,25=(x+1,5)^2-0,25\geq 0-0,25=-0,25$

Vậy $B_{\min}=-0,25$ khi $x=-1,5$

3.

$C=4x^2-8x=(4x^2-8x+4)-4=(2x-2)^2-4\geq 0-4=-4$

Vậy $C_{\min}=-4$ khi $2x-2=0\Leftrightarrow x=1$

4. Để $D_{\min}$ thì $5-x^2-2x$ là số thực âm lớn nhất

Mà không tồn tại số thực âm lớn nhất nên không tồn tại $x$ để $D_{\min}$

A\(=2x^2-8x+1\)

=2x(x-4)+1≥1

Min A=1 ⇔x=4

B=\(x^2+3x+2\)

\(=\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{1}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)≥\(-\dfrac{1}{4}\)

Min B=-1/4⇔x=-3/2

C=\(4x^2-8x\)

=\(\left(\left(2x\right)^2-2x.4+16\right)-16\)

=(2x-4)^2 -16≥-16

Min C=-16 ⇔x=2

D=\(\dfrac{1}{-\left(x^2-2x+1\right)+6}\)

=\(\dfrac{1}{-\left(x-1\right)^2+6}\)≥\(\dfrac{1}{6}\)

Min D=1/6 ⇔x=1