2 x 2 x 2 x 2= ??
gut gon nhak
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\(\left(x^2-2\right)\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=\left(x^2-x^3-2+2x\right)+\left(x^3+3^3\right)\)
\(=x^2-x^3-2+2x+x^3+27\)
\(=\left(-x^3+x^3\right)+x^2+2x+\left(-2+27\right)\)
\(=x^2+2x+25\)
\(A=\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow A=\left[\left(x^2+2-2x\right)\left(x^2+2+2x\right)\right]\left[\left(x^2-2\right)\left(x^2+2\right)\right]\)
\(\Leftrightarrow A=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(\Leftrightarrow A=\left(x^2+4x^2+4-4x^2\right)\left(x^4-4\right)\)
\(\Leftrightarrow A=\left(x^4+4\right)\left(x^4-4\right)\)
\(\Leftrightarrow A=x^{16}-16\)
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\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{\sqrt{x}+2}\left(dkxd:x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{x+4}\right)\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
Vậy \(B=\dfrac{1}{\sqrt{x}-2}\)
ta có B=2+\(\frac{x}{x-2}\)- \(\frac{4x^2}{x^2-4}\)- \(\frac{2-x}{x+2}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{2\left(x-2\right)\left(x+2\right)+x\left(x+2\right)-4x^2-\left(x-2\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left\{2\left(x-2\right)+x\right\}-\left\{4x^2-\left(x-2\right)^2\right\}}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(3x-4\right)-\left(2x-x+2\right)\left(2x+x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)\left(3x-4\right)-\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2\right)\left(3x-4-3x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-2}{x-2}\)
\(P=2\left(x^2-y^2\right)-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2\left(x^2-y^2\right)-4y^2+4xy\)
\(=2x^2-2y^2-4y^2+4xy\)
=2x^2+4xy-6y^2
2 x 2 x 2 x 2 = 4 x 2 x 2
= 8 x 2
= 16
mik mới khôi phục đc điểm âm mà giờ bị âm tiếp, các bạn giúp mik nha, giúp mik thì mik sẽ giúp lại
cảm ơn trc
2 x 2 x 2 x 2 = 2 x 4
= 8 k mik nhak cac bn