x³-2x²-5x=x(x²-2x-5)

\(x^3-2x^2-5x=x\left(x^2-2x-5\right)\)

Giúp vs m.n ơi mai mình kt òi

a) Với m=0

=> pt <=> \(x^2+5x=0\)

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

b) \(x^2+5x+3m=0\)

\(\Delta=25-12m\)

Để phương trình có 2 nghiệm phân biệt 

\(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow25-12m>0\)

\(\Leftrightarrow m< \dfrac{25}{12}\)

2x³ - 5x² - 3x = 0

⇔ x(2x² - 5x - 3) = 0

⇔ x(2x² + x - 6x - 3) = 0

⇔ x[(2x² + x) - (6x + 3)] = 0

⇔ x[x(2x + 1) - 3(2x + 1)] = 0

⇔ x(2x + 1)(x - 3) = 0

⇔ x = 0 hoặc 2x + 1 = 0 hoặc x - 3 = 0

*) 2x + 1 = 0

⇔ 2x = -1

⇔ x = -1/2

*) x - 3 = 0

⇔ x = 3

Vậy S = {-1/2; 0; 3}

Lời giải:

Đặt $(x-3)^2=a$. Khi đó pt đã cho tương đương với:

$(x^2-6x+9-9)^2+13(x-3)^2-77=0$

$\Leftrightarrow [(x-3)^2-9]^2+13(x-3)^2-77=0$

$\Leftrightarrow (a-9)^2+13a-77=0$

$\Leftrightarrow a^2-5a+4=0$

$\Leftrightarrow (a-1)(a-4)=0$

$\Leftrightarroe a=1$ hoặc $a=4$

Đến đây thì đơn giản rồi.

a) \(-10x^3+2x^2=0\)

\(\Rightarrow-2x^2\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(5x\left(x-2016\right)-x+2016=0\)

\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)

\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)

a: Ta có: \(-10x^3+2x^2=0\)

\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)

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

a. Bạn tự giải

b. 

Pt có 2 nghiệm phân biệt khi:

\(\Delta'=\left(m-3\right)^2-m^2>0\)

\(\Leftrightarrow-6m+9>0\)

\(\Leftrightarrow m< \dfrac{3}{2}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-3\right)\\x_1x_2=m^2\end{matrix}\right.\)

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

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

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

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

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

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

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

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

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

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

Vậy: \(S=\left\{1;3\right\}\)

\(d,x\left(2x-7\right)-4x+14=0\)

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

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

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

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

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

a) x² + 5x + 6

= x² + 2x + 3x + 6

= (x² + 2x) + (3x + 6)

= x(x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

b) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

c) x² - 5x - 14

= x² + 2x - 7x - 14

= (x² + 2x) - (7x + 14)

= x(x + 2) - 7(x + 2)

= (x + 2)(x - 7)

d) x² - 9x + 18

= x² - 3x - 6x + 18

= (x² - 3x) - (6x + 18)

= x(x - 3) - 6 (x - 3)

= (x - 3)(x - 6)

e) x² - 7x + 12

= x² -3x - 4x + 12

= (x² - 3x) - (4x + 12)

= x(x - 3) - 4(x - 3)

= (x - 3)(x - 4)

f) 3x² + 9x - 30

= 3(x² + 3x - 10)

= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)

= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)

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

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

 Chuc ban hoc tot

a) \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)

b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)

c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)

d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)

e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)