Ta có (x+1)² \(\ge0\)\(\ge\) với mọi x

=> 5(x+1)² \(\ge0\) với mọi

|y-3| \(\ge0\) với mọi y

=>5(x+1)²+|y-3| \(\ge0\) với mọi x,y

=>5(x+1)²+|y-3|-1 \(\ge-1\)

với mọi x,y

=> GTNN của biểu thức trên là -1 tại x=-1, y =3

Bài 1:

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2-26=0\)

\(\Leftrightarrow-13x-26=0\)

\(\Leftrightarrow-13\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

b) \(\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Bài 2:

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

b) \(\left(2x-1\right)\left(2x+1\right)\left(1-5x\right)\)

\(=\left(4x^2-1\right)\left(1-5x\right)\)

\(=4x^2-20x^3-1+5x\)

Bài 3: 

a: \(VT=\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3+x^2+x-x^2-x-1=x^3-1\)

b: \(x^4-y^4\)

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

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

c: \(x\left(2x-3\right)-2x\left(x+1\right)\)

\(=2x^2-3x-2x^2-2x=-5x⋮5\)

a) \(\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)

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

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

\(=\left(4x+y\right)y+xy\)

\(=\left[4\left(-2\right)+3\right].3+\left(-2\right).3\)

\(=\left(-8+3\right).3+1\)

\(=-15+1\)

\(=-14\)

thôi nha

a: \(A\ge-5\forall x,y\)

Dấu '=' xảy ra khi x=2 và y=-1

1) Tìm x và y biết

a) (2x+1)² + y² = 0

Ta có : \(\left(2x+1\right)^2\ge0;y^2\ge0\)

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

Để \(\left(2x+1\right)^2+y^2=0\)

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

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

b) x² + 2x + 1 + (y-1)² = 0

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

Lập luận tương tự câu a ,ta có :

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

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

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

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

c) x² - 2x + y² + 4y + 5 = 0

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)\)

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

Lập luận tương tự 2 câu trên

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

a, \(\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3\)

b, \(\left(a^3-2a^2+a-1\right)\left(a-5\right)\)

\(=a^4-2a^3+a^2-a-5a^3+10a^2-5a+5\)

\(=a^4-7a^3+11a^2-6a+5\)

c, \(\left(x^2-2x+y^2\right)\left(x-y\right)-3xy\left(y-x\right)\)

\(=x^3-2x^2+xy^2-x^2y+2xy-y^3-3xy^2+3x^2y\)

\(=x^3-2x^2-2xy^2-2x^2y+2xy-y^3\)

Bài 2:

a, \(\left(12x-5\right)\left(x+1\right)+\left(6x-2\right)\left(3-2x\right)=5\)

\(\Rightarrow12x^2+12x-5x-5+18x-12x^2-6+4x=5\)

\(\Rightarrow29x=5+5+6\)

\(\Rightarrow29x=16\Rightarrow x=\dfrac{16}{29}\)

b, \(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7=-8\)

\(\Rightarrow2x^2+3x-10x-15-2x^2+6x+x+7=-8\)

\(\Rightarrow0x=-8\Rightarrow x\in\varnothing\)

Chúc bạn học tốt!!!

Bài 1:

a) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

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

b) \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\cdot\left(4x^2-y^2\right)\)

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

Bài 2:

a) \(2x\cdot\left(x-5\right)-x\left(2x+3\right)=26\)

\(\Rightarrow2x^2-10x-2x^2-3x=26\)

\(\Rightarrow-13x=26\)

\(\Rightarrow x=2\)

b) \(\left(3y^2-y+1\right)\cdot\left(y-1\right)+y^2\cdot\left(4-3y\right)-\dfrac{5}{2}=0\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3-\dfrac{5}{2}=0\)

\(\Rightarrow2y+\dfrac{7}{5}=0\)

\(\Rightarrow2y=-1,4\)

\(\Rightarrow y=-0,7\)

c) \(2x^2+3\left(x-1\right)\cdot\left(x+1\right)=5x\left(x+1\right)\)

\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)

\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)

\(\Rightarrow5x^2-5x^2-5x=3\)

\(\Rightarrow-5x=3\)

\(\Rightarrow x=0,6\)

a, \(2x\left(x-5\right)-x\left(2x+3\right)=25\)

\(\Rightarrow2x^2-10x-2x^2-3x=25\)

\(\Rightarrow-13x=25\Rightarrow x=\dfrac{-25}{13}\)

b, \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\dfrac{5}{2}\)

\(\Rightarrow3y^3-y^2+y-3y^2+y-1+4y^2-3y^3=\dfrac{5}{2}\)

\(\Rightarrow2y-1=\dfrac{5}{2}\Rightarrow2y=\dfrac{7}{2}\Rightarrow y=\dfrac{7}{4}\)

c, \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

\(\Rightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)

\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)

\(\Rightarrow-5x=3\Rightarrow x=\dfrac{-3}{5}\)