a)xy(x²+2y)=xy.x²+xy.2y

                  =x³y+2xy²

b)-4(6x²-xy)=-4.6x²+4.xy

                   =-24x²+4xy

c)4x[x²+6x-1/2]

=4x.x²+4x.6x-4x.1/2

=4x³+24x²-2x

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

b) \(-4\times\left(6x^2-xy\right)=-24x^2+4xy\)

c)\(4x\times\left(x^2+6x-\frac{1}{2}\right)=4x^3+24x^2-2x\)

- Rút gọn đơn thức

a) 2x.(x^2 + 3x) - 2 (x^3 - 4 + 3x^2)

\(=2x^3+6x^2-2x^3+8-6x^2\)

\(=\left(2x^3-2x^3\right)+\left(6x^2-6x^2\right)-2x^3+8\)

\(=-2x^3+8\)

b) 2x.(5x + 2) - (10x^2 + 4x + 1)

Cách làm tương tự

\(A=x^4-x^2+2x+2020\)

\(A=\left(x^4-2x^2+1\right)+\left(x^2+2x+1\right)+2018\)

\(A=\left(x^2-1\right)^2+\left(x+1\right)^2+2018\)

\(A=\left(x-1\right)^2\left(x+1\right)^2+\left(x+1\right)^2+2018\)

\(A=\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]+2018\)

Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left[\left(x-1\right)^2+1\right]>0\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]\ge0\)

\(\Rightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]+2018\ge2018\)

\(\Rightarrow Amin=2018\Leftrightarrow\left(x+1\right)^2\left[\left(x-1\right)^2+1\right]=0\)

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

Mà \(\left(x-1\right)^2+1>0\) với mọi x

=> \(\left(x-1\right)^2+1\) vô nghiệm

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

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

Vậy GTNN của A là 2018 khi x = -1

Vậy GTNN của A là 2018 khi x

A=x²+2xy+2y²-2x-4y+2

=x²+xy-x+y²+xy-y-x-y+1+y²-2y+1

=(x²+xy-x)+(y²+xy-y)-(x+y-1)+(y²-2y+1)

= x(x+y-1)+y(y+x-1)-(x+y-1)+(y-1)²

=(x+y-1)(x+y-1)+(y-1)²

A=(x+y-1)²+(y-1)²

do (x+y-1)²\(\ge0\forall x;y\)

(y-1)²\(\ge0\forall y\)

=>(x+y-1)²+(y-1)²\(\ge0\)

=>Min A=0 khi

x+y-1=0

=>x+y=1 (*)

y-1=0

=>y=1

thay y=1 vào (*) ta đc

x+1=1

=>x=0

vậy....

3) \(B=3x^2+x+7\)

\(\Leftrightarrow B=3x^2+x+\dfrac{1}{12}+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left[x^2+2.x.\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2\right]+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)

Vậy GTNN của \(B=\dfrac{83}{12}\) khi \(x+\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{-1}{6}\)

B=3x² +x+7

=3x2+x+\(\dfrac{1}{12}+\dfrac{83}{12}\)

=3\(\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)

=3 \(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)

do \(\left(x+\dfrac{1}{6}\right)^2\ge0\forall x\)

=> \(3\left(x+\dfrac{1}{6}\right)^2\ge0\)

=> 3\(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\ge\dfrac{83}{12}\)

GTNN B=\(\dfrac{83}{12}\)

khi x+\(\dfrac{1}{6}\) =0

=>x=-\(\dfrac{1}{6}\)

a )\(2x\left(xy-3\right)+3xy\left(x+1-y\right)+3x\left(y^2-1\right)=2x^2y-6x+3x^2y+3xy-3xy^2+3xy^2-3x=5x^2y-9x+3xy\)

=> Phụ thuộc vào giá trị của biến

b) \(\left(x+2y\right)\left(x-2y\right)-x\left(x+4y^2\right)+5=x^2-4y^2-x^2-4xy^2+5=-4y^2-4xy^2+5\)

=> Phụ thuộc vào giá trị của biến

c) \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)=27x^3+8-9x^2+4=27x^3-9x^2+12\)

=> Phụ thuộc vào giá trị của biến

a: Ta có: \(2x\left(xy-3\right)+3xy\left(x-y+1\right)+3x\left(y^2-1\right)\)

\(=2x^2y-6x+3x^2y-3xy^2+3xy+3xy^2-3x\)

\(=5x^2y+3xy-9x\)

c: Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)\)

\(=27x^3+8-9x^2+4\)

\(=27x^3-9x^2+12\)

a)(x - 1) ^{x + 2} = (x - 1)^{x + 4}

=> (x - 1) ^{x + 4} - (x - 1)^{x + 2} = 0 

=> (x - 1)^{x + 2} . [(x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0^{x+2}\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)

Nếu x - 1 = 0

=> x = 1

Nếu x - 1 = - 1

=> x = 0

Nếu x - 1 = 1

=> x = 2

Vậy \(x\in\left\{0;1;2\right\}\)

b) \(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\)

\(\Rightarrow1,78^{2x-2}:1,78^x-1,78^x:1,78^x=0\)

\(\Rightarrow1,78^{x-2}-1=0\)

\(\Rightarrow1,78^{x-2}=1\)

\(\Rightarrow1,78^{x-2}=1,78^0\)

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

Vậy x = 2

Thanks you bạn nhìu😉😉😉😉😉😉

\(A=-\frac{1}{2}\left(4x^2+y^2+4+4xy-8x-4y\right)-\frac{3}{2}y^2-4\)

\(A=-\frac{1}{2}\left(2x+y-2\right)^2-\frac{3}{2}y^2-4\le-4\)

\(\Rightarrow A_{max}=-4\) khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(A_{min}\) không tồn tại