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6x2 + 7x - 3
= 6x2 + 9x - 2x - 3
= (6x2 - 2x) + ( 9x - 3)
= 2x (3x - 1) + 3 (3x - 1)
= (3x - 1) (2x + 3)
t i c k nha!! 3645745756785685874363464455675687678996348658
6x2 + 7x - 3
= 6x2 + 9x - 2x - 3
= 3x.(2x + 3) - (2x + 3)
= (2x + 3).(3x - 1)
\(2x^3-3x^2+3x-1=x^3+x^3-3x^2+3x-1\)
=\(x^3+\left(x^3-3x^2+3x-1\right)\)=\(x^3+\left(x-1\right)^3\)
=\(\left(x+x-1\right)\left(x^2-x\left(x-1\right)+\left(x-1\right)^2\right)\)
=\(\left(2x-1\right)\left(x^2-x^2+x+x^2-2x+1\right)\)
=\(\left(2x-1\right)\left(x^2-x+1\right)\)
dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)+64x^4\)
=\(\left[\left(x^2-2x+2\right)^4+2.10x^2\left(x^2-2x+2\right)^2+100x^4\right]\)-100x4+64x2
=\(\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^2\)
=\(\left[\left(x^2-2x+2\right)^2+4x^2\right].\left[\left(x^2-2x+2\right)^2+16x^2\right]\)
phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung(không dùng HĐT)
x3 -3x2y+3xy2 - y3
\(=xy\left(x^2-3x+3y-y^2\right)\)
\(=xy\left[\left(x-y\right)\left(x+y\right)+3\left(x-y\right)\right]\)
\(=xy\left(x-y\right)\left(x+y+3\right)\)
\(Ht\)
nếu sai cho mik xl vì mik chx thành thục cái này
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
a) (2x - 1)2 - (x + 3)2
= (2x - 1 - x - 3).(2x - 1 + x + 3)
= (x - 4).(3x + 2)
b) x2.(x - 3) + 12 - 4x
= x2.(x - 3) - 4x + 12
= x2.(x - 3) - 4.(x - 3)
= (x - 3).(x2 - 4)
= (x - 3).(x - 2).(x + 2)
Áp dụng HĐT:
a2 - b2 = (a - b)(a + b)
\(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
2x2-3x-2
=2x2-4x+x-2
=2x.(x-2)+(x-2)
=(2x+1).(x-2)