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3x^2 - 8x + 4
= 3x^2 - 6x + 2x + 4
= 3x(x - 2) + 2(x - 2)
= (x - 2) (3x + 2)
\(A=x^3-x^2-8x+12\)
\(=x^3-2x^2+x^2-2x-6x+12\)
hay \(A=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+6\right)\)
\(=\left(x+2\right)^2\left(x+3\right)\)
\(A=x^3-x^2-8x+12\)
\(=x^3-2x^2+x^2-2x-6x+12\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-6\right)\)
\(=\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(x-2\right)^2\left(x+3\right)\)
Chúc bạn học tốt.
Mình nhầm 1 chút nhé mọi người \(x^2-8x-9\) nhé!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a) \(x^2-6x+8\)
\(=x^2-2\cdot x\cdot3+3^2-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
Còn lại tương tự
a) \(x^2-6x+8=x^2-2x-4x+8\)
\(=\left(x^2-2x\right)-\left(4x-8\right)\)
=x(x-2)-4(x-2) = (x-2)(x-4)
\(x^2+7x+12\)
cách 1: \(=x^2+4x+3x+12\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
cách 2: \(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
cách 3: \(=\left(x^2+7x+12,25\right)-0.25\)
\(=\left(x+3.5\right)^2-0.5^2\)
\(=\left(x+3.5+0.5\right)\left(x+3.5-0.5\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
lấy đâu ra 8 cách vậy trời!!!!!!!!!!!!!!!
Cách 1:
\(x^2+7x+12\)
\(=\left(x^2+4x\right)+\left(3x+12\right)\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
Ta có: \(-8x^2+23x+3\)
\(=\left(-8x^2+24x\right)-\left(x-3\right)\)
\(=-8x\left(x-3\right)-\left(x-3\right)\)
\(=\left(-8x-1\right)\left(x-3\right)\)
\(=\left(3-x\right)\left(8x+1\right)\)
\(-8x^2+23x+3\)
\(=-\left(8x^2-23x-3\right)\)
\(=-\left(8x^2-24x+x-3\right)\)
\(=-\left[8x\left(x-3\right)+\left(x-3\right)\right]\)
\(=-\left(8x+1\right)\left(x-3\right)\)
a) Đặt \(x^2=y\Rightarrow x^4+x^2-20=y^2+y-20=y^2-4y+5y-20=\left(y-4\right)\left(y+5\right)\)
Thay trở lại, ta có: \(x^4+x^2-20=\left(x^2-4\right)\left(x^2+5\right)=\left(x-2\right)\left(x+2\right)\left(x^2+5\right)\)
b) Đặt \(x-y=z\Rightarrow\left(x-y\right)^2+4x-4y-12=z^2+4z-12=z^2-2z+6z-12=\left(z-2\right)\left(z+6\right)\)
Thay trở lại ta có kết quả sau: \(\left(x-y-2\right)\left(x-y+6\right)\)
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5 \left(x-3\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
\(x^2-8x+15\)
\(=\left(x^2-3x\right)-\left(5x-15\right)\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
Tham khảo nhé~
\(=\left(x^2y+xy^2\right)-\left(x+y\right)=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
C1
Ta có \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-6\right)\left(x-2\right)\)
C2
ta có \(x^2-8x+12\)
\(=x^2-8x+16-4\)
\(=\left(x-4\right)^2-2^2\)
\(=\left(x-4-2\right)\left(x-4+2\right)\)
\(=\left(x-6\right)\left(x-2\right)\)