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Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
2x3 + 6x2 = x2 + 3x
⇔ (2x3 + 6x2) – (x2 + 3x) = 0
⇔ 2x2(x + 3) – x(x + 3) = 0
⇔ x(x + 3)(2x – 1) = 0
(Nhân tử chung là x(x + 3))
⇔ x = 0 hoặc x + 3 = 0 hoặc 2x – 1 = 0
+ x + 3 = 0 ⇔ x = -3.
+ 2x – 1 = 0 ⇔ 2x = 1 ⇔ x = 1/2.
Vậy tập nghiệm của phương trình là 
`2x^3 +6x^2 =x^2 +3x`
`<=> 2x^3 +6x^2 -x^2 -3x=0`
`<=> 2x^3 +5x^2 -3x=0`
`<=> x(2x^2 +5x-3)=0`
`<=> x(2x^2 +6x-x-3)=0`
`<=> x[2x(x+3)-(x+3)]=0`
`<=> x(2x-1)(x+3)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x-1=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
b)
`(2+x)^2 -(2x-5)^2=0`
`<=> (2+x-2x+5)(2+x+2x-5)=0`
`<=> (-x+7)(3x-3)=0`
\(< =>\left[{}\begin{matrix}-x+7=0\\3x-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
`a) 2x^3 + 6x^2 = x^2 + 3x`
`=> 2x^3 + 6x^2 - x^2 - 3x = 0`
`=> 2x^3 + 5x^2 - 3x = 0`
`=> x(2x^2 + 5x - 3) = 0`
`=> x (2x^2 + 6x - x - 3) = 0`
`=> x [(2x^2 + 6x) - (x+3)] = 0`
`=> x [2x(x+3) - (x+3)] = 0`
`=> x (2x - 1)(x+3) = 0`
`=> x = 0` hoặc `2x - 1 = 0` hoặc `x + 3 = 0`
`=> x = 0` hoặc `x = 1/2` hoặc `x = -3`
`b) (2+x)^2 - (2x-5)^2 = 0`
`=> (2+x+2x-5)(2+x-2x+5) = 0`
`=> (3x - 3)(7-x) = 0`
`=> 3x - 3 = 0` hoặc `7 - x = 0`
`=> x = 1` hoặc `x = 7`
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
⇔ (3x – 1)(x2 + 2) – (3x – 1)(7x – 10) = 0
⇔ (3x – 1)(x2 + 2 – 7x + 10) = 0
⇔ (3x – 1)(x2 – 7x + 12) = 0
⇔ (3x – 1)(x2 – 4x – 3x + 12) = 0
⇔ (3x – 1)[(x2 – 4x) – (3x - 12)] = 0
⇔ (3x – 1)[x(x – 4) – 3(x – 4)] = 0
⇔ (3x – 1)(x – 3)(x – 4) = 0
⇔ 3x – 1 = 0 hoặc x – 3 = 0 hoặc x – 4 = 0
+ 3x – 1 = 0 ⇔ 3x = 1 ⇔ x = 1/3.
+ x – 3 = 0 ⇔ x = 3.
+ x – 4 = 0 ⇔ x = 4.
Vậy phương trình có tập nghiệm là 