a: \(F=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)

\(=-\left(2x-y\right)^2\cdot\left[2x-y+x\right]-y^3\)

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

\(=\left(-4x^2+4xy-y^2\right)\left(3x-y\right)-y^3\)

\(=-12x^3+4x^2y+12x^2y-4xy^2-3xy^2+y^3-y^3\)

\(=-12x^3+16x^2y-7xy^2\)

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

mà \(\left(x-2\right)^2+y^2>=0\forall x,y\)

nên dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)

=>x=2 và y=0

Thay x=2 và y=0 vào F, ta được:

\(F=-12\cdot2^3+16\cdot2^2\cdot0-7\cdot2\cdot0^2\)

\(=-12\cdot2^3\)

\(=-12\cdot8=-96\)

b: \(G=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=x^3+y^3+3\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)

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

\(=x^3+y^3+24x^3-3y^3\)

\(=25x^3-2y^3\)

Ta có: \(\left\{{}\begin{matrix}x+y=2\\y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-3\\x=2-y=2-\left(-3\right)=2+3=5\end{matrix}\right.\)

Thay x=5 và y=-3 vào G, ta được:

\(G=25\cdot5^3-2\cdot\left(-3\right)^3\)

\(=25\cdot125-2\cdot\left(-27\right)\)

\(=3125+54=3179\)

c: \(H=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3-26y^3\)

Ta có: \(\left\{{}\begin{matrix}3x-y=5\\x=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=3x-5=3\cdot2-5=1\end{matrix}\right.\)

Thay x=2 và y=1 vào H, ta được:

\(H=28\cdot2^3-26\cdot1^3\)

\(=28\cdot8-26\)

=198

Bài 2: Tính giá trị của biểu thức sau:

\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)

Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)

\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)

Bài 4: Tìm x

a) \(9x^2+x=0\)

\(\Rightarrow x\left(9x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)

b) \(27x^3+x=0\)

\(\Rightarrow x\left(27x^2+1=0\right)\)

\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)

Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)

Vậy \(x=0\)

Bài 1:

\(B=\dfrac{1}{9}x^2-2x+9\)

\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)

\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)

\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

MINH CUNG VAY

A = 5x(x - y) - y(5x - y)

A = 5x² - 5xy - 5xy + y²

A = 5x² - 10xy + y² (1)

Thay x = -1; y = 3 vào (1), ta có:

5.(-1)² - 10.(-1).3 + 3² = 44

B = 4y(x² - 3xy + 3y²) - 2xy(2x - 6y - 3)

B = 4x²y - 12x² + 12y³ - 4x²y + 12xy² + 6xy

B = 12y³ + 6xy (1)

Thay x = 5; y = -1 vào (1), ta có:

12.(-1)³ + 6.5.(-1) = -42

C = 5x²(x - y²) + 3x(xy² - y) - 5x³ 

C = 5x³ - 5x²y² + 3x²y² - 3xy - 5x³ 

C = -2x²y² - 3xy (1)

Thay x = -2; y = -5 vào (1), ta có:

-2.(-2)².(-5)² - 3.(-2).(-5) = -230

D = 6x²(y² - xy + 2x²y) - 3xy(2xy - x² + 4x³)

D = 6x²y² - 6x³y + 12x⁴y - 6x²y² + 3x³y - 12x⁴y

D = -3x³y (1)

Thay x = 11; y = -1 vào (1), ta có:

-3.11³.(-1) = 3993

\(D=50^2-49.51\)

\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)

\(\Leftrightarrow D=50^2-50^2+1=1\)

\(C=39^2+78.61+61^2\)

\(\Leftrightarrow C=39^2+2.39.61+61^2\)

\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)

a) (7x + 4) _²  - (7x + 4)(7x - 4) 

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

\(=56x+32\)

A = (3x + y)^2 - 3y . ( 2x - 1/3y )

=2y²+9x²

B = ( x - 2 )^2 + ( x + 2 )^2 - 2. ( x - 2 ) ( x + 2)

=2⁴

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

=y³+x³

D = ( x -5 ) ( x+ 5 ) -(x - 8 ) (x + 4)

=4x+7

E = (3x + 1 )^2 - 2 . ( 9x^2 - 1 ) + ( 3x - 1 ) ^2

=4

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

=6x-18