7,x²-2xy+y²+3x-3y=(x-y)²+3(x-y)=(x-y)(x-y+3)

8,x⁴+4=(x⁴+4x²+4)-4x²=(x²+2)²-(2x)²=(x²-2x+2)(x²+2x+2)

9,4x(x+1)²-5x²(x+1)-4.(x+1)=(x+1)\(\left[4x\left(x+1\right)-5x^2-4\right]\)=(x+1)(4x²+4x-5x²-4)=(x+1)(-x²+4x-4)=-(x+1)(x-2)²

7: \(x^2-2xy+y^2+3x-3y\)

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

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

8: \(x^4+4\)

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

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

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

9: \(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right)\)

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

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

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

10: \(\left(1+2x\right)\left(1-2x\right)-\left(x+2\right)\left(x-2\right)\)

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

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

\(=-5\left(x-1\right)\left(x+1\right)\)

46:

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

\(=3x^2-2x+1-3x^2+7x-2=5x-1\)

Khi x=-0,2 thì A=-1-1=-2

45:

a: \(=\dfrac{-5x^6}{3x^2}=-\dfrac{5}{3}x^4\)

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

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

.

a, 36:(x–5) =  2 2

(x–5) = 9

x = 14

b, [3.(70–x)+5]:2 = 46

[3.(70–x)+5] = 92

70–x = 29

x = 41

c, 450:[41–(2x–5)] =  3 2 .5

41–(2x–5) = 10

2x–5 = 31

2x = 36

x = 18

d, 230+[ 2 4 +(x–5)] = 315. 2018 0

16+(x–5) = 315–230

x–5 = 85–16

x = 69+5

x = 74

e,  2 x + 2 x + 1  = 48

2 x .(2+1) = 48

2 x = 16 =  2 4

x = 4

f,  3 x + 2 + 3 x  = 2430

3 x . 3 2 + 1 = 2430

3 x = 2430:10 = 243 =  3 5

x = 5

a,x=0;-2

b,..de ma bn

a/ x² .(x -2)=0

suy ra hoặc x²=0

hoặc x-2=0

hoặc cả hai =0

từ đó tính ra x

Từ \(2x=5y\) \(\Rightarrow\) \(\dfrac{x}{5}=\dfrac{y}{2}\)

Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

\(\Rightarrow3x+4y=3\cdot5k+4\cdot2k=15k+8k=23k=46\)

\(\Rightarrow k=46:23=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\cdot2=10\\y=2\cdot2=4\end{matrix}\right.\)