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x^7+x^5+1=x^7+x^6+x^5-x^6+1
=x^5(x^2+x+1)-[(x^3)^2-1]
=x^5(x^2+x+1)-(x^3+1)(x^3-1)
=x^5(x^2+x+1)-(x^3+1)(x-1)(x^2+x+1)
=(x^2+x+1)[x^5-(x^3+1)(x-1)]
=(x^2+x+1)(x^5-x^4+x^3-x+1)
\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)
ta có (x-1)(x-2)(x-3)(x-4)-15=(x-1)(x-4)(x-2)(x-3)-15=\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15\)(*)
đặt \(t=x^2-5x+5\)thì pt (*) =\(\left(t-1\right)\left(t+1\right)-15=t^2-1-15\)\(=t^2-16=\left(t+4\right)\left(t-4\right)=\)\(\left(x^2-5x+5+4\right)\left(x^2-5x+5-4\right)=\)\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)\)
Bài khó quá
4x⁴+4x-3= (2x)²+2(2x)+1-4
=(2x+1)²-2²=(2x+1-2)(2x+1+2)
=(2x-1)(2x+3)
x^3-x+6=x^3+2x^2-2x^2-4x+3x+6=x^2.(x+2)-2x.(x+2)+3.(x+2)=(x^2-2x+3).(x+2)
\(4x-x^3+2x^2y-xy^2\)
\(=x\left(4-x^2+2xy-y^2\right)\)
\(=x\left(4-\left(x^2-2xy+y^2\right)\right)\)
\(=x\left(2^2-\left(x-y\right)^2\right)\)
\(=x\left(2+x-y\right)\left(2-x+y\right)\)
\(\left(x-y\right)^3-x^3+y^3=\left(x-y\right)^3-\left(x^3-y^3\right)=\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)=-3xy\left(x-y\right)\)
\(\left(x-y\right)^3-x^3+y^3\\ =\left(x-y\right)^3-\left(x^3-y^3\right)\\ =\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)\\ =\left(x-y\right)\left[\left(x-y\right)^2-\left(x^2+xy+y^2\right)\right]\\ =\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)\\ =\left(-3xy\right)\left(x-y\right)\)
de sai ak
Đề đúng đó bạn