rut gon \(^{x^2}\)-x+2x-2
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\(2x^2\left(x-2\right)-2x\left(x-1\right)\left(x+1\right)=2x^3-4x^2-2x^3+2x=-4x^2+2x=-2x\left(2x-1\right)\)
\(2x^2\left(x-2\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=2x^3-4x^2-2x\left(x^2-1\right)\)
\(=2x^3-4x^2-2x^3+2x=-4x^2+2x\)
\(A=\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow A=\left[\left(x^2+2-2x\right)\left(x^2+2+2x\right)\right]\left[\left(x^2-2\right)\left(x^2+2\right)\right]\)
\(\Leftrightarrow A=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(\Leftrightarrow A=\left(x^2+4x^2+4-4x^2\right)\left(x^4-4\right)\)
\(\Leftrightarrow A=\left(x^4+4\right)\left(x^4-4\right)\)
\(\Leftrightarrow A=x^{16}-16\)
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\(=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}:\frac{2\left(x+1\right)^2}{4\left(x-1\right)^2}=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}.\frac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=2\)
Cái này giống dạng toán thường thôi bạn ạ! Để mình giải cho bạn nhé!
x-2/x-1 + 2x(1-x)/x^2-x = x-2/x-1 + (-2x(x-1))/x(x-1) = x-2/x-1 + (-2x/x) = x^2-2x/x(x-1) +( -2x(x-1)/x(x-1)) = x^2-2x+(-2x(x-1))/x(x-1)
= x^2-2x-2x^2+2x/x(x-1) = -x^2/x(x-1) = -x/x-1
\(A=\left(2x-2\right)^2-4x\left(x-2\right)+46\)
\(\Leftrightarrow A=4x^2-8x+4-4x^2+8x+46\)
\(\Leftrightarrow A=50\)
2ax^2-a(1+2x^2)-[a-x(x+a)]
=2ax2-2ax2+a+x2-ax+a
=(2ax2-2ax2)-(a+a)+ax+x2
=0-2a+ax+x2
=x2+ax-2a
\(x^2-x+2x-2=\)\(\left(x^2-x\right)+\left(2x-2\right)\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
\(x^2-x+2x-2=\left(x^2-x\right)+\left(2x-2\right)\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)