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c1: x2-10x+24=(x2-2.x.5+52)-1=(x-5)2-1=(x-5-1)(x-5+1)=(x-6)(x-4)
c2: x2-10x+24=x2-4x-6x+24=x(x-4)-6(x-4)=(x-6)(x-4)
c1 =x2-6x-4x+24
=x(x-6)-4(x-6)
=(x-4)(x-6)
c2 =x2-10x+25-1
=(x-5)2-1
=(x-5-1)(x-5+1)
=(x-6)(x-4)
= (x^4-4x^3)+(3x^3-12x^2)+(2x^2-8x)-(2x-8)
= x^3.(x-4)+3x^2.(x-4)+2x.(x-4)-2.(x-4)
= (x-4).(x^3+3x^2+2x-2)
Tk mk nha
x3-7x2+10x
=x3-2x2-5x2+10x
=(x3-2x2)-(5x2-10x)
=x2(x-2)-5x(x-2)
=(x2-5x)(x-2)
=x(x-5)(x-2)
giải phương trình:
\(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(x^2-10x+16\)
\(=\left(x^2-2x\right)-\left(8x-16\right)\)
\(=x.\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x-8\right)\)
Tham khảo nhé~
\(x^4+5x^3+10x-4\)
\(=x^4+5x^3-2x^2+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
Mình cũng vừa làm được cách 2:
\(x^4+5x^3+10x-4\)
=\(x^4-4+5x^3+10x\)
=\(\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
=\(\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2+4x-3x-12\right)\)
\(=\left(x-2\right)\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
Ta có:\(x^3-x^2-14x+24=\left(x^3-2x^2\right)+\left(x^2-2x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2-3x+4x-12\right)\)
\(=\left(x-2\right)\left[x\left(x-3\right)+4\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
x^2 - 10x + 24
= x^2 - 4x - 6x + 24
= x(x - 4) - 6(x - 4)
= (x - 6)(x - 4)
ko vt lại đề
x2-6x-4x+24
=(x2-6x)-(4x-24)
=x(x-6)-4(x-6)
=(x-6)(x-4)