mình vừa lên lớp 9 , chưa học phương trình bậc 2 

a)2x³ + 7x² - x - 12 =0

=>2x³+x²-4x+6x²+3x-12=0

=>x(2x²+x-4)+3(2x²+x-4)=0

=>(x+3)(2x²+x-4)=0

=>x+3=0 hoặc 2x²+x-4=0

Xét x+3=0 <=>x=-3

Xét 2x²+x-4=0 ta dùng delta

\(\Delta=1^2-\left(-4\left(2.4\right)\right)=33>0\)

=>pt có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{33}}{4}\)

b)- x^3 + x^2 + 7x + 2 =0

=>-x³+3x²+x-2x²+6x+2=0

=>-x(x²-3x-1)+(-2)(x²-3x-1)=0

=>-(x+2)(x²-3x-1)=0

=>-(x+2)=0 hoặc x²-3x-1=0

Xét -(x+2)=0 <=>x=-2

Xét x²-3x-1=0 theo delta ta có:

\(\Delta=\left(-3\right)^2-\left(-4\left(1.1\right)\right)=13>0\)

=>pt cũng có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\)

xài hóc ne đi

a)

`x^2 +5x+6=0`

`<=> x^2 + 3x +2x+6=0`

`<=> x(x+3)+2(x+3)=0`

`<=> (x+3)(x+2)=0`

`<=> x+3=0 hoặcx+2=0`

`<=> x=-3 hoặc x=-2`

b)

`x^2 -7x+6=0`

`<=> x^2 -6x-x+6=0`

`<=> x(x-6)-(x-6)=0`

`<=> (x-6)(x-1)=0`

`<=> x-6=0 hoặc x-1=0 `

`<=> x=6 hoặc x=1`

c)

`x^2 +x -12=0`

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

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

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

`<=> x+4=0 hoặc x-3=0`

`<=> x=-4 hoặc x=3`

d)

`x^2 -x-6=0`

`<=>x^2 -3x+2x-6=0`

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

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

`<=> x-3=0 hoặc x+2=0`

`<=> x=3 hoặc x=-2`

e)

`2x^2 -3x-5=0`

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

`<=> x(2x-5)+(2x-5)=0`

`<=> (2x-5)(x+1)=0`

`<=> 2x-5=0 hoặc x+1=0`

`<=> x=5/2 hoặc x=-1`

a)\(\left\{{}\begin{matrix}3x-2y=3\\2x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\3x-2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\3-2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b)\(x^2+7x+12=0\)

\(\Leftrightarrow x^2+3x+4x+12=0\)( chị nghĩ + 12 đúng hơn á )

\(\Leftrightarrow x\left(x+3\right)+4\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)

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

a)Ta có \(\left(2x+1\right)\left(x^2+2\right)=0\)<=>

2x+1=0<=>x=\(-\frac{1}{2}\)

hoặc \(x^2+2=0\)<=>\(x^2=-2\)(Vô lí)

Vậy tập nghiệm của pt S=(\(-\frac{1}{2}\))

b)\(\left(x^2+4\right)\left(7x-3\right)=0\)

<=>\(\left[{}\begin{matrix}x^2+4=0\\7x-3=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x^2=-4\\x=\frac{3}{7}\end{matrix}\right.\)

\(x^2=-4\) vô lí

Vậy ..........

c)\(\left(x^2+x+1\right)\left(6-2x\right)=0\)

<=>\(\left[{}\begin{matrix}x^2+x+1=0\\6-2x=0\end{matrix}\right.\)

Vì \(x^2+x+1>0\)(dễ dàng c/m)

=>6-2x=0=>x=3

Vậy...

d)\(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

<=>8x-4=0,x=\(\frac{1}{2}\)

hoặc \(x^2+2x+2=0\)(vô lí)

Vậy .....

\(1;x^2+7x+10=0\Rightarrow x^2+2x+5x+10=0\Rightarrow x\left(x+2\right)+5\left(x+2\right)=0\)

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

=> x + 2 = 0 hoặc x + 5 = 0

=> x = -2 hoặc x = - 5

2, x^4 - 5x^2 +  4 = 0 

x^4  - 4x^2  - x^2 + 4 = 0 

x^2 ( x^2 - 4) - ( x^2 - 4) = 0 

( x^2 - 1)( x^2 - 4) = 0 

( x - 1 )( x + 1)( x - 2)( x + 2) = 0

=> x= 1 hoặc x= -1 hoặc x = 2 hoặc x = - 2

Đúng cho mi8nhf mình giải tiếp cho

(4x - 3)² - (2x + 1)² = 0

\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

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

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

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

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

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

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)