ta có 4 x 3 y 2   –   8 x 2 y 3   =   4 x 2 y 2 . x   –   4 x 2 y 2 . 2 y   =   4 x 2 y 2 ( x   –   2 y )    

Vậy 4x³y² – 8x²y³ = 4x²y²(x – 2y)      

Đáp án cần chọn là: C

bấm đúng cho mik đi 

Bài 2:

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

\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)

\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)

\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)

Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)

Bài 1:

a)x²-4x²y+4xy

=x(x-4xy+y)

b)đề sai

Bài 3:

3yx + 6x - y = 7

<=> x(3y+6) - (3y+6) = 27

<=> (3y+6)(x+1) = 27

Ta có bảng sau:

Vậy...

nhanh nhanh ho minh voi

4x = 3y => \(\frac{x}{3}=\frac{y}{4}\) 

3y = 6z => \(\frac{y}{6}=\frac{z}{3}\) => \(\frac{y}{4}=\frac{z}{2}\)

=> \(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\) => \(\frac{x}{3}=\frac{2y}{8}=\frac{3z}{6}\)

Áp dung dãy tỉ số bằng nhau ta có

      \(\frac{x}{3}=\frac{2y}{8}=\frac{3z}{6}=\frac{x-2y+3z}{3-8+6}=\frac{5}{1}=5\)

=> x = 15; y = 20 và z = 10

\(4x=3y=6z\)

\(\frac{4x}{12}=\frac{3y}{12}=\frac{6z}{12}\)

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\)

\(\frac{x}{3}=\frac{2y}{8}=\frac{3z}{6}=\frac{x-2y+3z}{3-8+6}=\frac{5}{1}=5\)

\(\begin{cases}x=15\\y=20\\z=10\end{cases}\)

\(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\left(1\right)\)

\(5y=3z\Rightarrow\frac{y}{3}=\frac{z}{5}\left(2\right)\)

Từ (1) (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\) và 2x - 3y + z = 6

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{9.2+12.2+20}=\frac{6}{2}=3\)

=> \(\frac{x}{9}=3\) => x = 9.3 = 27

=> \(\frac{z}{20}=3\) => z = 60

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

giúp mình nhé

a)\(x^2+2y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)

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

\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)

Vậy x=-1 y=1

a)  \(x^2+2y^2+2xy-2y+1=0\)

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

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

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

b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)

\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)

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

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

            \(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)