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\(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-3^2=\left(x^2+2-3\right)\left(x^2+2+3\right)=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2\right)^2-3^2\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
nhớ nha
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)
\(=\left(x-y+2\right)^2-9\)
\(=\left(x-y+2-3\right)\left(x-y+2+3\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
\(x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2-1+4\left(x-y-1\right)\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+4\right)\)
\(=\left(x-y-1\right)\left(x-y+5\right)\)
\(x^5-5x^3+4x=x\left(x^4-5x^2+4\right)=x\left(x^4-x^2-4x^2+4\right)\)
\(=x\left[x^2\left(x^2-1\right)-4\left(x^2-1\right)\right]=x\left(x^2-1\right)\left(x^2-4\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(x^5-5x^3+4x=x^5-4x^3-x^3+\) \(4x\)
\(=\) \(x^3.\left(x^2-4\right)-x.\left(x^2-4\right)\)
\(=\left(x^3-x\right)\left(x^2-4\right)\)
\(=x\left(x^2-1\right)\left(x^2-4\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
Đề sai rồi bạn ạ thật đó
phương thảo: mình giải đc r, đề k sai đâu