x²-4x+3

x²-x-3x-3

=(x²-x)-(3x-3)

=x(x-1) - 3(x-1)

=(x-3)(x-1)

C1:       \(x^2-4x+3\)

\(=x^2-4x+4-1\)

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

\(=\left(x-2-1\right).\left(x-2+1\right)\)

\(=\left(x-3\right).\left(x-1\right)\)

C2 :    \(x^2-4x+3\)

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

\(=x.\left(x-1\right)-3.\left(x-1\right)\)

\(=\left(x-1\right).\left(x-3\right)\)

(x-1)(x-3)

4x²-4x-15=4x²-10x+6x-15=2x(2x-5)+3(2x-5)=(2x+3)(2x-5)

ths                                           .

\(4x^2-4x-15\)

\(=4x^2-10x+6x-15\)

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

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

a) x² + 4x + 3

= x² + 3x + x +3

= ( x² + 3 ) + ( x + 3 )

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

= ( x + 3 ) ( x + 1 )

b) 4x² - 4x - 3

= 4x² + 2x - 6x - 3

= ( 4x² + 2x ) - ( 6x + 3 )

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

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

c) x² - x - 12

= x² + 3x - 4x - 12

= ( x² + 3x ) - ( 4x + 12 )

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

= ( x + 3 ) ( x - 4 )

d) 4x⁴ - 4x²y² - 8y⁴

= 4 ( x⁴ - x²y² - 2y⁴ )

Hk tốt

cảm ơn bạn

a) \(x^2+4x+3\)

= \(x^2+x+3x+3\)

= \(x\left(x+1\right)+3\left(x+1\right)\)

= \(\left(x+1\right)\left(x+3\right)\)

b) \(4x^2-4x-3\)

= \(4x^2+2x-6x-3\)

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

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

c)\(x^2-x-12\)

= \(x^2-4x+3x-12\)

= \(x\left(x-4\right)+3\left(x-4\right)\)

= \(\left(x-4\right)\left(x+3\right)\)

câu d la 4x^2.y^2 phai ko

\(4x^4-4x^2y^2-8y^4\)

= \(4x^4-8x^2y^2+4x^2y^2-8y^4\)

= \(4x^2\left(x^2-2y^2\right)+4y^2\left(x^2-2y^2\right)\)

= \(\left(4x^2+4y^2\right)\left(x^2-2y^2\right)\)

=\(4\left(x^2+y^2\right)\left(x^2-2y^2\right)\)

Kết bạn với mình nha

Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

\(=2x\left(2x+9\right)-\left(2x+9\right)\)

\(=\left(2x-1\right)\left(2x+9\right)\)

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

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

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

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

ấn máy tính để tìm nghiệm rồi phân tích ra

\(x^3-4x^2+4x-1\)

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

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

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

này đinh đức hùng không phân tích nữa à , cái đằng sau còn phân tích được nữa đấy

\(a,x^2-5x+4=x^2-4x-x+4=x\left(x-4\right)-\left(x-4\right)=\left(x-4\right)\left(x-1\right)\)

\(b,4x^2-4x-3=4x^2-2.2x.1+1-3-1=\left(2x-1\right)^2-4=\left(2x-1-2\right)\left(2x-1+2\right)=\left(2x-3\right)\left(2x+1\right)\)

\(3\left(4x-5\right)^2-8x+10=0\)

\(\Leftrightarrow\)\(3\left(4x-5\right)\left(4x-5\right)-2\left(4x-5\right)=0\)

\(\Leftrightarrow\)\(\left(4x-5\right)\left[2\left(4x-5\right)-2\right]=0\)

\(\Leftrightarrow\)\(\left(4x-5\right)\left(8x-10-2\right)=0\)

\(\Leftrightarrow\)\(\left(4x-5\right)\left(8x-12\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}4x-5=0\\8x-12=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{4}\\x=1,5\end{cases}}\)

Vậy...

k mk đi

ai k mk 

mk k lại 

thanks

a/ \(x^2-4x+3=\left(x^2-x\right)-\left(3x-3\right)=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

b/ \(3x^2-5x+2=\left(3x^2-3x\right)-\left(2x-2\right)=3x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(3x-2\right)\)

\(x^2-4x+3\)

\(=x^2-3x-x-3\)

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

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

\(3x^2-5x+2\)

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

\(=3x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(3x-2\right)\left(x-1\right)\)

a/ x²-4x+3=x²-x-3x+3=x(x-1)-3(x-1)=(x-3)(x-1)

b/ 3x²-5x+2=3x²-3x-2x+2=3x(x-1)-2(x-1)=(3x-2)(x-1)