Bài 1:

a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)

b: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)

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

bài 2:

a: Đề thiếu vế phải rồi bạn

b: \(x^3-13x=0\)

=>\(x\left(x^2-13\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)

Bài 1:

a, $3x-6y$

$=3(x-2y)$

b, $14x^2y-21xy^2+28x^2y^2$

$=7xy(2x-3y+4xy)$

c, $10x(x-y)-8y(y-x)$

$=10x(x-y)-8y[-(x-y)]$

$=10x(x-y)+8y(x-y)$

$=(x-y)(10x+8y)$

$=2(x-y)(5x+4y)$

Bài 2:

a, Đề thiếu rồi bạn nhé.

b, \(x^3-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)

2.

a. 3x(12x - 4) - 9x(4x - 3) = 30

<=> 36x² - 12x - 36x² + 27x = 30

<=> 36x² - 36x² - 12x + 27x = 30

<=> 15x = 30

<=> x = 2

b. x(5 - 2x) + 2x(x - 1) = 15

<=> 5x - 2x² + 2x² - 2x = 15

<=> -2x² + 2x² + 5x - 2x = 15

<=> 3x = 15

<=> x = 5

a) x² ( 5x³ - x - 2323x²y=  6969x³y²- 2323x⁴y+ 2323x²y²

c) x² ( 4x^{3 -} 5xy + 2x ) ( -

a)-(x-y)(x²+xy-1)=-(x³+x²y-x-x²y-xy²+y)

                          =-(x³-xy²-x+y)

                          =-x³+xy²+x-y

b)x²(x-1)-(x³+1)(x-y)=x³-x²-x³+x²y-x+y

                                =-x²+x²y-x+y

c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x²-3x-4x+2-15x²-10x-3x-2

                                             =-9x²-20x

d) hình như bạn ghi lỗi

Bài 2: C=x(x²-y)-x²(x+y)+y(x²-x)

             =x³-xy-x³-x²y+x²y-xy

             =-2xy

Thay x=1/2,y=-1 vào C, ta có:

        C=-2.1/2.(-1)=1

Vậy C=1 khi x=1/2 và y=-1.

`@` `\text {Ans}`

`\downarrow`

`a)`

`5x^3 - x - 1/2`

Đã thu gọn?

`b)`

`(3xy - x^2 + y) * 2/3x^2y`

`= 3xy * 2/3 x^2y - x^2* 2/3x^2y + y*2/3x^2y`

`= 2x^3y^2 - 2/3x^4y + 2/3x^2y^2`

`c)`

`(4x^3 - 5xy +2x) (-1/2xy)`

`= 4x^3* (-1/2xy) - 5xy* (-1/2xy) + 2x * (-1/2xy)`

`= -2x^4y + 5/2x^2y^2 - x^2y`

`d)`

`(x^2 - 2x +1) (x-1)`

`= x^2(x-1) - 2x(x-1) + x - 1`

`= x^3 - x^2 - 2x^2 + 2x + x -1`

`= x^3 -3x^2 + 3x - 1`

d: =(x-1)^3=x^3-3x^2+3x-1

c: =-2x^4y+5/2x^2y^2-x^2y

a.

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

b.

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

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

Từ giả thiết bài toán suy ra

y ≥ 0 x 2 2 ≤ - 2 x 2 + 3 x ⇔ y ≥ 0 5 x 2 - 6 x ≤ 0 ⇔ y ≥ 0 0 ≤ x ≤ 6 5  

Ta có

  x 2 + y 2 ≤ x 2 + - 2 x 2 + 3 x 2 = 4 x 4 - 12 x 3 + 10 x 2

Ta có  f ' x = 4 x x - 1 x - 5

f ' x = 0 x = 0 x = 1 x = 5  So điều kiện, chọn x = 0 ; x = 1 ; f(0); f(1) = 2;  f 6 5 = 1224 625

Vậy m a x P = 2

Đáp án D