\(A=-3x^2+6x-7=-3\left(x^2-2x+1-1\right)-7\)

\(=-3\left(x-1\right)^2-4\le-4\)Dấu ''='' xảy ra khi x = 1

\(B=-2x^2+5x+1=-2\left(x^2-\dfrac{5}{2}x\right)+1\)

\(=-2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}-\dfrac{25}{16}\right)+1\)

\(=-2\left(x-\dfrac{5}{4}\right)^2+\dfrac{33}{8}\le\dfrac{33}{8}\)Dấu ''='' xảy ra khi x = 5/4

C;D chỉ có GTNN thôi bạn nhé \(C=2x^2-8x+13=2\left(x^2-4x+4-4\right)+13\)

\(=2\left(x-2\right)^2+5\ge5\)Dấu ''='' xảy ra khi x = 2

\(D=x^2-3x+5=x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+5\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)Dấu ''='' xảy ra khi x = 3/2 

d: Ta có: \(D=x^2-3x+5\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

lớp 8 có pt bậc 2 ak??

Có nhưng giải bằng PT tích nhé

a: \(P=x^2+y^2-6x-2y+17\)

\(=x^2-6x+9+y^2-2y+1+7\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-1=0

=>x=3 và y=1

b: \(Q=x^2+xy+y^2-3x-3y+999\)

\(=x^2+x\left(y-3\right)+y^2-3y+999\)

\(=x^2+2\cdot x\cdot\left(\frac12y-\frac32\right)+\left(\frac12y-\frac32\right)^2+y^2-3y-\left(\frac12y-\frac32\right)^2+999\)

\(=\left(x+\frac12y-\frac32\right)^2+y^2-3y-\left(\frac14y^2-\frac32y+\frac94\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34y^2-\frac32y-\frac94+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y-3\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y+1-4\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y-1\right)^2+996\ge996\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x+\frac12y-\frac32=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=-\frac12y+\frac32=-\frac12+\frac32=\frac22=1\end{cases}\)

c: \(R=2x^2+2xy_{}+y^2-2x+2y+15\)

\(=x^2-4x+4+x^2+2xy+y^2+2x+2y+11\)

\(=\left(x-2\right)^2+x^2+2xy+y^2+2x+2y+1+10\)

\(=\left(x-2\right)^2+\left(x+y+1\right)^2+10\ge10\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y+1=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x-1=-2-1=-3\end{cases}\)

d: \(S=x^2+26y^2-10xy+14x-76y+59\)

\(=x^2-10xy+25y^2+14x-70y+y^2-6y+59\)

\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+y^2-6y+9+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-3=0\\ x-5y+7=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=5y-7=5\cdot3-7=15-7=8\end{cases}\)

e: \(T=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x-2y+5=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=2y-5=2\cdot1-5=2-5=-3\end{cases}\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a:

ĐKXĐ: \(x^2+3x>=0\)

=>x(x+3)>=0

=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)

 \(\sqrt{16}-\sqrt{x^2+3x}=0\)

=>\(\sqrt{x^2+3x}=\sqrt{16}\)

=>x^2+3x=16

=>x^2+3x-16=0

\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)

Do đó: Phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)

b:

ĐKXĐ: \(x\in R\)

 \(3x-1-\sqrt{4x^2-12x+9}=0\)

=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)

=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)

c:

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)

 \(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)

\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)

=>\(x^2-4x+3=0\)

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

=>x=3(loại) hoặc x=1(nhận)

\(x\left(5-6x\right)+\left(2x-1\right)\left(3x+\text{4}\right)=6\\ \Leftrightarrow5x-6x^2+6x^2+8x-3x-4=6\)

\(\Leftrightarrow10x-4=6\)

\(\Leftrightarrow10x=6+4\\ \Leftrightarrow10x=10\\ \Leftrightarrow x=\dfrac{10}{10}\)

\(\Leftrightarrow x=1\)

\(x^2\left(x-2021\right)-x+2021=0\)

\(\Leftrightarrow x^2\left(x-2021\right)-(x-2021)=0\)

\(\Leftrightarrow\left(x-2021\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-2021\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2021=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=1\\x=-1\end{matrix}\right.\)

Câu 1: \(x^4-8x^2-9=0\)

=>\(x^4-9x^2+x^2-9=0\)

=>\(\left(x^2-9\right)\left(x^2+1\right)=0\)

=>\(x^2-9=0\)

=>\(x^2=9\)

=>x=3 hoặc x=-3

=>Tập nghiệm là S={3;-3}

a) \(x^3-x^2+3x-3>0\)

\(\Leftrightarrow x^2\left(x-1\right)+3\left(x-1\right)>0\)

\(\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\) 

Mà: \(x^2+3>0\forall x\) 

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

b) \(x^3+x^2+9x+9< 0\)

\(\Leftrightarrow x^2\left(x+1\right)+9\left(x+1\right)< 0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+1\right)< 0\)

Mà: \(x^2+9>0\forall x\)

\(\Leftrightarrow x+1< 0\)

\(\Leftrightarrow x< -1\)

d) \(4x^3-14x^2+6x-21< 0\)

\(\Leftrightarrow2x^2\left(2x-7\right)+3\left(2x-7\right)< 0\)

\(\Leftrightarrow\left(2x^2+3\right)\left(2x-7\right)< 0\)

Mà: \(2x^2+3>0\forall x\)

\(\Leftrightarrow2x-7< 0\)

\(\Leftrightarrow2x< 7\)

\(\Leftrightarrow x< \dfrac{7}{2}\)

d) \(x^2\left(2x^2+3\right)+2x^2>-3\)

\(\Leftrightarrow2x^4+3x^2+2x^2+3>0\)

\(\Leftrightarrow2x^4+5x^2+3>0\)

\(\Leftrightarrow\left(x^2+1\right)\left(2x^2+3\right)>0\) 

Mà: 

\(x^2+1>0\forall x\)

\(2x^2+3>0\forall x\)

\(\Rightarrow x\in R\)

a: =>x^2(x-1)+3(x-1)>0

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

=>x-1>0

=>x>1

b: =>x^2(x+1)+9(x+1)<0

=>(x+1)(x^2+9)<0

=>x+1<0

=>x<-1

c: 4x^3-14x^2+6x-21<0

=>2x^2(2x-7)+3(2x-7)<0

=>2x-7<0

=>x<7/2

d: =>x^2(2x^2+3)+2x^2+3>0

=>(2x^2+3)(x^2+1)>0(luôn đúng)

a: =x^2-3x+9/4+11/4

=(x-3/2)^2+11/4>=11/4>0 với mọi x

b: =x^2-x+1/4+11/4

=(x-1/2)^2+11/4>0

c: =2(x^2-x+5/2)

=2(x^2-x+1/4+9/4)

=2(x-1/2)^2+9/2>=9/2>0

d: =3(x^2-2/3x+1/3)

=3(x^2-2/3x+1/9+2/9)

=3(x-1/3)^2+2/3>=2/3>0 với mọi x