\(a,\Rightarrow\left(2x-5\right)^2+2\left(2x-5\right)\left(x+2\right)+\left(x+2\right)^2=0\\ \Rightarrow\left(2x-5+x+2\right)^2=0\\ \Rightarrow3x-3=0\\ \Rightarrow x=1\\ b,\Rightarrow9-\left(x^2-5x\right)^2=9\\ \Rightarrow x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a) x³y + x - y - 1

= (x³y - y) + (x - 1)

= y(x³ - 1) + (x - 1)

= y(x - 1)(x² + x + 1) + (x - 1)

= (x - 1)[y(x² + x + 1) + 1]

= (x - 1)(x²y + xy + y + 1)

b) x²(x - 2) + 4(2 - x)

= x²(x - 2) - 4(x - 2)

= (x - 2)(x² - 4)

= (x - 2)(x - 2)(x + 2)

= (x - 2)²(x + 2)

c) x³ - x² - 20x

= x(x² - x - 20)

= x(x² + 4x - 5x - 20)

= x[(x² + 4x) - (5x + 20)]

= x[x(x + 4) - 5(x + 4)]

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

d) (x² + 1)² - (x + 1)²

= (x² + 1 - x - 1)(x² + 1 + x + 1)

= (x² - x)(x² + x + 2)

= x(x - 1)(x² + x + 2)

e) 6x² - 7x + 2

= 6x² - 3x - 4x + 2

= (6x² - 3x) - (4x - 2)

= 3x(2x - 1) - 2(2x - 1)

= (2x - 1)(3x - 2)

f) x⁴ + 8x² + 12

= x⁴ + 2x² + 6x² + 12

= (x⁴ + 2x²) + (6x² + 12)

= x²(x² + 2) + 6(x² + 2)

= (x² + 2)(x² + 6)

g) (x³ + x + 1)(x³ + x) - 2

Đặt u = x³ + x

x³ + x + 1 = u + 1

(u + 1).u - 2

= u² + u - 2

= u² - u + 2u - 2

= (u² - u) + (2u - 2)

= u(u - 1) + 2(u - 1)

= (u - 1)(u + 2)

= (x³ + x - 1)(x³ + x + 2)

= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)

= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]

= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]

= (x³ + x - 1)(x - 1)(x² - x + 2)

h) (x + 1)(x + 2)(x + 3)(x + 4) - 1

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1

= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)

Đặt u = x² + 5x + 4

u + 2 = x² + 5x + 6

(1) u.(u + 2) - 1

= u² + 2u - 1

= u² + 2u + 1 - 2

= (u² + 2u + 1) - 2

= (u + 1)² - 2

= (u + 1 + √2)(u + 1 - √2)

= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)

= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)

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

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

`a,x^2 +4x-5=0`

`<=> x^2-x+5x-5=0`

`<=> x(x-1)+5(x-1)=0`

`<=>(x-1)(x+5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)

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

`b, x^2 -x-12=0`

`<=> x^2 +3x-4x-12=0`

`<=>(x^2+3x)-(4x+12)=0`

`<=>x(x+3)-4(x+3)=0`

`<=>(x+3)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\)

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

`c, (2x-7)^2 - 6(2x-7)(x-3)=0`

`<=>(2x-7)(2x-7 -6x+18)=0`

`<=>(2x-7) ( -4x+11)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\-4x+11=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\-4x=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{matrix}\right.\)

a: =>(x+5)(x-1)=0

=>x=1 hoặc x=-5

b: =>(x-4)(x+3)=0

=>x=4 hoặc x=-3

c: =>(2x-7)(2x-7-6x+18)=0

=>(2x-7)(-4x+11)=0

=>x=11/4 hoặc x=7/2

tham khảo: https://hoc24.vn/cau-hoi/.2256230161739

a) ⇔ \(4x^2+4x-x-1=0\)

⇔ \(4x^2+3x-1=0\)

⇔ \(4x(x+1)-(x+1)=0\)

⇔ \((x+1)(4x-1)=0\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)

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b) \(x^3-4x^2+4x=0\)

⇔ \(x^2(x-2)-2x(x-2)=0\)

⇔ \((x-2)(x^2-2x)=0\)

⇒ \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

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c) \(x^2-3x+2=0\)

⇔ \(x(x-2)-(x-2)=0\)

⇔ \((x-1)(x-2)=0\)

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

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