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1/ đề sai vd: 2+3=5 là số nguyên tố
2/ \(4x^2-a^2+y^2-16b^2+4xy+8ab\)
\(=\left[\left(2x\right)^2+2.2xy+y^2\right]-\left[a^2+2.4ab-\left(4b\right)^2\right]\)
\(=\left(2x+y\right)^2-\left(a-4b\right)^2\)
\(=\left(2x+y+a-4b\right)\left(2x+y-a+4b\right)\)
3/
\(M=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+5x-x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-5^2\)
\(=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2+4x\right)^2\ge0\)
\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\)
\(\Rightarrow M\ge-25\)
Dấu "=" xảy ra khi x = 0 hoặc x = -4
Vậy Mmin = -25 khi x = 0 hoặc x = -4
xm+2+xm = xm(x^2 + 1)
xx+1-xx-1 = x^x . x - x^x - \(\frac{x^x}{x^x}\)=x^x ( x - 1 - 1/x^x) ( cái này mik ko bik đúng hay sai)
a^3 - 1 = a^3 - 1^3 = ( a-1)(a^2 + a + 1)
a^5 - b^5 = \(\left(\sqrt{a^5}\right)^2-\left(\sqrt{b^5}\right)^2=\left(\sqrt{a^5}+\sqrt{b^5}\right)\left(\sqrt{a^5}-\sqrt{b^5}\right)\)
a,\(-4x^2+4x-1\)
\(\Leftrightarrow\left(-2x-1\right)^2\)
b,\(\left(2x+1\right)^2-4\left(x-1\right)^2\)
\(\Rightarrow\left[2x+1-2\left(x-1\right)\right].\left[2x+1+2\left(x-1\right)\right]\)
\(\Rightarrow\left(2x+1-2x+2\right)\left(2x+1+2x-2\right)\)
\(\Rightarrow3\left(4x-1\right)\)
c,\(\left(2x-y\right)^2-4x^2+12x-9\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(4x^2-12x+9\right)\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(2x-3\right)^2\)
\(\Leftrightarrow\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)
\(\Rightarrow\left(y+3\right)\left(4x+y-3\right)\)
d,\(\left(x+1\right)^2-4\left(x+1\right)y^2+4y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+2^2y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+4\left(y^2\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)-2y^2+\left(2y^2\right)^2\)
\(\Leftrightarrow\left(x+1-2y^2\right)^2\)
\(x^2\left(x-2\right)^2-\left(x-2\right)^2-x^2+1\)
\(=\left(x-2\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-2\right)^2-1\right]\left(x^2-1\right)\)
Làm tiếp cho chắc nhé
\(=\left(x-2-1\right)\left(x-2+1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-3\right)\left(x-1\right)^2\left(x+1\right)\)