d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

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

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

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

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

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

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a)\(7x\left(x-2\right)=\left(x-2\right)\)

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

\(\Leftrightarrow\left(7x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7x-1=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=2\end{matrix}\right.\)

b)\(4x^2-9-x\left(2x-3\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)

c)\(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+9x+5x^2+45=0\)

\(\Leftrightarrow x\left(x^2+9\right)+5\left(x^2+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+9\right)=0\)

Dễ thấy: \(x^2+9\ge 9 >0\forall x\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

d,e tương tự

1) \(x^3+5x^2+9x=-45\)

\(\Rightarrow x^2\left(x+5\right)+9x+45=0\)

\(\Rightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Rightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2=-9\left(loai\right)\\x=-5\left(nhan\right)\end{cases}}\)

2) \(x^3-6x^2-x+30=0\)

\(\Rightarrow x^3-3x^2-3x^2+9x-10x+30=0\)

\(\Rightarrow x^2\left(x-3\right)-3x\left(x-3\right)-10\left(x-3\right)=0\)

\(\Rightarrow\left(x^2-3x-10\right)\left(x-3\right)=0\)

\(\Rightarrow\left(x^2-5x+2x-10\right)\left(x-3\right)=0\)

\(\Rightarrow\left[x\left(x-5\right)+2\left(x-5\right)\right]\left(x-3\right)=0\)

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

\(\)Từ đây giải x giống câu trên nhé.

3) \(x^2+16=10x\)

\(\Rightarrow x^2-10x+16=0\)

\(\Rightarrow\left(x-8\right)\left(x-2\right)=0\)

Tương tự....

a) (x + 3)² - (x - 2)² = 2x

=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x

=> 5(2x + 1) = 2x

=> 10x + 5 = 2x

=> 10x - 2x = -5

=> 8x = -5

=> x = -5/8

b) 7x(x - 2) = x - 2

=> 7x(x - 2) - (x - 2) = 0

=> (7x - 1)(x - 2) = 0

=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)

c) 8x³ - 12x² + 6x - 1 = 0

=> (2x - 1)³ = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = 1/2

Lời giải:

a) $8x^3-12x^2+6x-1=0$

$\Leftrightarrow (2x)^3-3(2x)^2.1+3.2x.1^2-1^3=0$

$\Leftrightarrow (2x-1)^3=0$

$\Leftrightarrow 2x-1=0\Leftrightarrow x=\frac{1}{2}$
b)

\(4x^2-9-x(2x-3)=0\)

$\Leftrightarrow (2x-3)(2x+3)-x(2x-3)=0$

$\Leftrightarrow (2x-3)(2x+3-x)=0$

$\Leftrightarrow (2x-3)(x+3)=0$

$\Rightarrow 2x-3=0$ hoặc $x+3=0$

$\Leftrightarrow x=\frac{3}{2}$ hoặc $x=-3$

c)

\(x^3+5x^2+9x=-45\)

\(\Leftrightarrow (x^3+5x^2)+(9x+45)=0\)

$\Leftrightarrow x^2(x+5)+9(x+5)=0$

$\Leftrightarrow (x+5)(x^2+9)=0$

Vì $x^2+9>0$ với mọi $x$ nên $x+5=0\Leftrightarrow x=-5$

d)

$x^3-6x^2-x+30=0$

$\Leftrightarrow x^3-3x^2-3x^2+9x-10x+30=0$

$\Leftrightarrow x^2(x-3)-3x(x-3)-10(x-3)=0$

$\Leftrightarrow (x-3)(x^2-3x-10)=0$

$\Leftrightarrow (x-3)(x^2+2x-5x-10)=0$

$\Leftrightarrow (x-3)[x(x+2)-5(x+2)]=0$

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

$\Rightarrow x=3; x=-2$ hoặc $x=5$

g)

$x^2+16=10x$

$\Leftrightarrow x^2-10x+16=0$

$\Leftrightarrow x^2-10x+25-9=0$

$\Leftrightarrow (x-5)^2-3^2=0\Leftrightarrow (x-5-3)(x-5+3)=0$

$\Leftrightarrow (x-8)(x-2)=0$

$\Rightarrow x=8$ hoặc $x=2$

a) x³ - 7x + 6

= x³ - 2x² + 2x² - 4x - 3x + 6

= x² ( x - 2 ) + 2x ( x - 2 ) - 3 ( x - 2 )

= ( x - 2 ) ( x² + 2x - 3 )

= ( x - 2 ) ( x² - x + 3x - 3 )

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

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

b ) x³ - 9x² + 6x + 16

= x³ - 8x² - x² + 8x - 2x + 16

= x² ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )

= ( x - 8 ) ( x² - x - 2 )

= ( x - 8 ) ( x² + x - 2x - 2 )

= ( x - 8 ) [ x ( x + 1 ) - 2 ( x + 1 ) ]

= ( x - 8 ) ( x + 1 ) ( x - 2 )

c ) x³ - 6x² - x + 30

= x³ - 5x² - x² + 5x - 6x + 30

= x² ( x - 5 ) - x ( x - 5 ) - 6 ( x - 5 )

= ( x - 5 ) ( x² - x - 6 )

= ( x - 5 ) ( x² - 3x + 2x - 6 )

= ( x - 5 ) [ x ( x - 3 ) + 2 ( x - 3 ) ]

= ( x - 5 ) ( x - 3 ) ( x + 2 )

d ) 2x³ - x² + 5x + 3

= 2x³ + x² - 2x² - x + 6x + 3

= x² ( 2x + 1 ) - x ( 2x + 1 ) + 3 ( 2x + 1 )

= ( 2x + 1 ) ( x² - x + 3 )

Dài 166

b) 2x²+3x-27=2x²-6x+9x-27=2x(x-3)+9(x-3)=(x-3)(2x+9)