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

Nhớ cho 5 sao luôn nhé

Ta có: \(4x^2-8x+7=4x^2-8x+4+3\left(2x-2\right)^2+3\ge3\)

\(\Rightarrow B>0\)

Vậy B có GTLN \(\Leftrightarrow\left(2x-2\right)^2+3\)có GTNN

Mà \(\left(2x-2\right)^2+3\ge3\Rightarrow Min\left(4x^2=8x+7\right)=3\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

\(\Rightarrow\)Max B = 3\(\Leftrightarrow x=1\)

toán 12 nha

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 :)) 

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

\(minA=-3\Leftrightarrow x=2\)

b) \(B=-x^2-8x+5=-\left(x+4\right)^2+21\le21\)

\(maxB=21\Leftrightarrow x=-4\)

c) \(C=2x^2-8x+19=2\left(x-2\right)^2+11\ge11\)

\(minC=11\Leftrightarrow x=2\)

d) \(D=-3x^2-6x+1=-3\left(x+1\right)^2+4\le4\)

\(maxD=4\Leftrightarrow x=-1\)

a) A = (x-2)^2 - 3 >= -3

--> A nhỏ nhất bằng -3

 <=> x = 2

b) B = -(x+4)^2 + 21 <= 21

--> B lớn nhất bằng 21

<=> x = -4