a)\(2x^3+3x^2+2x+3=0\)

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

\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\x^2+1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=-3\\x^2+1>0\left(loai\right)\end{array}\right.\)

\(\Leftrightarrow x=-\frac{3}{2}\)

b)\(x\left(2x-1\right)\left(1-2x\right)=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(2x-1\right)^2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\2x=1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)

\(2x^3+3x^2+2x+3=0\)

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

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

\(2x+3=0\left(x^2+1\ge1>0\right)\)

\(2x=-3\)

\(x=-\frac{3}{2}\)

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

\(\left[\begin{array}{nghiempt}x=0\\2x-1=0\\1-2x=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\2x=1\\2x=1\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)

???giải ròi mà:

x²+5x-6

= x²-x+6x-6

= x(x-1)+6(x-1)

= (x-1).(x+6)

a) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ac\)

\(=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2ac\)

\(=b^2-2bc+2ac=b.\left(b-2c+2a\right)\)

b) \(x^4+2x^3+5x^2+4x-12\)

\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)

\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\)

\(=\left(x-1\right)\left[x^2.\left(x+2\right)+x.\left(x+2\right)+6.\left(x+2\right)\right]\)

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

Pạn Khánh Châu ơi

Cái dòng thứ 2 đấy, dấu hiệu nhận biết là j vậy

Mà sao pạn phân tích hay vậy????

Dễ nhưng mà dài chết người 

giải dùm mình với đi ạ,mình cảm ơn

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

thanks

Bài 3

a) 2x(x - 3) - x + 3 = 0

2x(x - 3) - (x - 3) = 0

(x - 3)(2x - 1) = 0

x - 3 = 0 hoặc 2x - 1 = 0

*) x - 3 = 0

x = 3

*) 2x - 1 = 0

2x = 1

x = 1/2

Vậy x = 1/2; x = 3

b) (3x - 1)(2x + 1) - (x + 1)² = 5x²

6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0

(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1

-x = 2

x = -2

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

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

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

x³-5x²+x-5=0

=> x².(x-5)+(x-5)=0

=> (x-5).(x²+1)=0

=> x-5=0                 hoặc x²+1=0

=> x=5                    hoặc x²=-1 (vô lí)

Vậy x=5.

x⁴-2x³+10x²-20x=0

=> x³.(x-2)+10x(x-2)=0

=> (x-2).(x³+10x)=0

=> x.(x-2).(x²+10)=0

=> x=0      hoặc x-2=0               hoặc x²+10=0

=> x=0       hoặc x=2                 hoặc x²=-10 (vô lí)

Vậy x=0 hoặc x=2.

x=0,x=2

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).