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\(\times^2+7\times+12\)
\(=(\times^2+4\times)+\left(3\times+12\right)\)
\(=\times\left(\times+4\right)+3\left(\times+4\right)\)
\(=\left(\times+4\right)\left(\times+3\right)\)
\(x^2+7x+12=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
Ta có : x4 + 8x2 + 7x + 8
= x4 - x + 8x2 + 8x + 8
= x(x3 - 1) + 8(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 - x)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 8)
Học tốt nhé !
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
\(x^2-7x+12\)
\(=\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
\(x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
\(x^2+6x+8\)
\(=x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)=\left(x+2\right)\left(x+4\right)\)
a) x2 + 7x + 12
= x2 + 3x + 4x + 12
= x.(x+3) + 4.(x+3)
= (x+3).(x+4)
b) x2 + 6x + 8
= x2 + 2x + 4x + 8
= x.(x+2) + 4.(x+2)
= (x+2).(x+4)
\(x^2-7x+9\)
\(=x^2-2\cdot x\cdot\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{13}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2\)
\(=\left(x-\frac{7}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{7}{2}+\frac{\sqrt{13}}{2}\right)\)
\(=\left(x-\frac{7+\sqrt{13}}{2}\right)\left(x-\frac{7-\sqrt{13}}{2}\right)\)
Ta có:\(x^3-7x-6=\left(x^3-3x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)=\left(x-3\right)\left(x^2+2x+x+2\right)\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
=x3-x-6x-6
=(x3-x)-(6x-6)
=x(x2-1)-6(x-1)
=x(x-1)(x+1)-6(x-1)
=(x-1)(x2+1-6)
Gửi Thắng Nguyễn: Mình không biết tại sao lại ko phân tích được?
\(=2x^3+2x^2-9x^2-9x+10x+10\)
\(=2x^2\left(x+1\right)-9x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2-9x+10\right)\)
\(=\left(x+1\right)\left[\left(2x^2-4x\right)-\left(5x-10\right)\right]\)
\(=\left(x+1\right)\left[2x\left(x-2\right)-5\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(2x-5\right)\)
\(x^2+7x+12\)
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:
\(x^2+7x+12\)
\(=\left(x^2+4x\right)+\left(3x+12\right)\)
\(=x\left(x+4\right)+3\left(x+4\right)\)
\(=\left(x+3\right)\left(x+4\right)\)