a) \(2x^2+2x+1=0\)

\(\Rightarrow2x^2+2x=-1\)

\(\Rightarrow2x\left(x+1\right)=-1\)

⇒ Pt vô nghiệm

a: \(2x^2+2x+1=0\)

\(\text{Δ}=2^2-4\cdot2\cdot1=4-8=-4< 0\)

Vì Δ<0 nên phương trình vô nghiệm

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy \(S=\left\{2;2\right\}\)

A= 2x^2 + y^2 - 2xy -2x+3

A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2

A= (x-y)^2 + (x-1)^2 + 2

(x-y)^2> hoặc = 0 với mọi giá trị của x

(x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2

=> A lớn hơn hoặc bằng 2

=> GTNN của A=2 tại x=y=1

Yêu cầu đề là gì vậy bạn?

Câu 2: 

a: \(\Leftrightarrow3x^2+2x-1=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow x^3-4x-x^3-8=4\)

hay x=-3

a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)

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

=2x^2-3x+4

b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)

\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)

c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

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

\(a,\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{7}{4}=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\\ \Leftrightarrow x,y\in\varnothing\left[\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}>0\right]\\ b,\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\\ \Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\\ \Leftrightarrow x,y,z\in\varnothing\left[\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\right]\)

\(c,\Leftrightarrow-\left(x^2-10xy+25y^2\right)-\left(y^2-20y+100\right)-50=0\\ \Leftrightarrow-\left(x-5y\right)^2-\left(y-10\right)^2-50=0\\ \Leftrightarrow x,y\in\varnothing\left[-\left(x-5y\right)^2-\left(y-10\right)^2-50\le-50< 0\right]\)