\(6x-1=2x+3\\ \Rightarrow4x=4\\ \Rightarrow x=1\)

\(5x\left(x-2\right)+\left(2x^4+10x^3-4x^2\right):x^2\\ =5x^2-10x+2x^2\left(2x^2+5x-4\right):x^2\\ =5x^2-10x-2\left(2x^2+5x-4\right)\\ =5x^2-10x-4x^2-10x+8\\ =x^2-20x+4\)

\(\left(x+2\right)^2-2x-4=0\\ \Rightarrow\left(x+2\right)^2-2\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(x+2-2\right)=0\\ \Rightarrow x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

a)

`4(x-2)^2 =4`

`<=>(x-2)^2 =1`

`<=>x-2=1` hoặc `x-2=-1`

`<=>x=3` hoặc `x=1`

b)

`5(x^2 -6x+9)=5`

`<=>(x-3)^2 =1`

`<=>x-3=1`hoặc `x-3=-1`

`<=>x=4` hoặc `x=2`

c)

`4x^2 +4x+1=0`

`<=>(2x+1)^2 =0`

`<=>2x+1=0`

`<=>x=-1/2`

d)

`9x^2 +6x+1=2`

`<=>(3x+1)^2 =2`

\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)

\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

1. (x² + 3)(x⁴ - 3x² + 9)

= x⁶ + 27

2. (2x + 1)(4x² - 2x + 1)

= 8x³ + 1

3. (x² + 2)(x⁴ - 2x² + 4)

= x⁶ + 8

4. (3x + 2)(9x² - 6x + 4)

= 27x³ + 8

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

2: \(\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)

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

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

b)x²-2x+1=4

⇔(x-1)²=4

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

c)x²-4x+4=9

⇔ (x-2)²=9

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

d)4x²-4x+1=4

⇔ (2x-1)²=4

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

e)x²-2x-8=0

⇔ x²-4x+2x-8=0

⇔ x(x-4)+2(x-4)=0

⇔(x-4)(x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

f)9x²-6x-8=0

⇔ 9x²-12x+6x-8=0

⇔ 3x(3x-4)+2(3x-4)=0

⇔ (3x-4)(3x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3