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\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

a)

\(512-\left(128-5x\right)=3x+12\\ 512-128+5x=3x+12\\ 384+5x=3x+12\\ 5x-3x=12-384\\ 2x=-372\\ x=\left(-372\right):2\\ x=-186\)

b)

\(\left(2x-1\right)+\left(4x-2\right)+...+\left(400x-200\right)=5+10+...+1000\\ \left(2x+4x+...+400x\right)-\left(1+2+...+200\right)=5+10+...+1000\\ x\left(2+4+...+400\right)=\left(5+10+...+1000\right)+\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=5\cdot\left(1+2+...+200\right)+1\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=\left(5+1\right)\cdot\left(1+2+...+200\right)\\ 2x\cdot\left(1+2+...+200\right)=6\cdot\left(1+2+...+200\right)\\ \Rightarrow2x=6\\ x=6:2\\ x=3\)

c)

\(\left(x+2\right)+\left(4x+4\right)+\left(7x+6\right)+...+\left(25x+18\right)+\left(28x+20\right)=1560\\ \left(x+4x+7x+...+25x+28x\right)+\left(2+4+6+...+18+20\right)=1560\\ x\left(1+4+7+...+25+28\right)+110=1560\\ 145x+110=1560\\ 145x=1560-110\\ 145x=1450\\ x=1450:145\\ x=10\)

d)

\(x+4x+5x+9x+14x+...+97x=500\\ x\left(1+4+5+9+14+...+97\right)=500\)

Dãy số trong ngoặc có quy luật: Số thứ \(n\) bằng số thứ \(n-1\) cộng số thứ \(n-2\)

Suy ra dãy số đó là: \(1+4+5+9+14+23+37+60+97=250\)

Thế vào ta được:

\(250x=500\\ x=500:250\\ x=2\)

e)

\(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\\ 720-41+\left(2x-5\right)=8\cdot5\\ 720-41+2x-5=40\\ \left(720-41-5\right)+2x=40\\ 674+2x=40\\ 2x=40-674\\ 2x=-634\\ x=\left(-634\right):2\\ x=-317\)

f)

\(697:\dfrac{15x+364}{x}=17\\ \dfrac{15x+364}{x}=697:17\\ \dfrac{15x+364}{x}=41\\ \dfrac{15x+364}{x}\cdot x=41x\\ 15x+364=41x\\ 364=41x-15x\\ 364=26x\\ x=364:26\\ x=14\)

a, \(14x^2-21x^2+28x^2y^2\)

\(=-7x^2+28x^2y^2=-7x^2\left(1-4y^2\right)\)

\(=-7x\left(1-2y\right)\left(1+2y\right)\)

b, \(8x^3+4x^2-y^3-y^2\)

\(=\left(8x^3-y^3\right)+\left(4x^2-y^2\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x-y\right)\left(2x+y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2+2x+y\right)\)

\(a=\sqrt{2}+\sqrt{7-2\sqrt{5}-1}+1\)

\(=\sqrt{2}+\sqrt{5}-1+1=\sqrt{2}+\sqrt{5}\)

f(x)=x^4(x+2)-14x^2(x+2)+9(x+2)+1

=(x+2)(x^4-14x^2+9)+1

\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left[\left(7+2\sqrt{10}\right)^2-14\left(7+2\sqrt{10}\right)+1\right]\)+1

\(=\left(\sqrt{2}+\sqrt{5}+2\right)\left(89+28\sqrt{10}-84-28\sqrt{10}+1\right)\)+1

=6(căn 2+căn 5+1)+1

\(=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

Bài 1:

a) \(2x\left(x^2-5x+6\right)=2x^3-10x^2+12x\)

b) \(\left(7x^5+14x^2y^3-28x^3y^2\right):7x^2=x^3+2y^3-4xy^2\)

Bài 2:

\(x^2+y^2+2x-8y+17=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

a) \(a^2+2ab+b^2-2a-2b+1=\left(a+b\right)^2-2\left(a+b\right)+1\)

\(=\left(a+b\right)\left(a+b\right)-2\left(a+b\right)+1=\left(a+b-2\right)\left(a+b\right)+1\)

b) \(x^3-4x^2+4x-1=x^3-3x^2+x-x^2+3x-1\)

\(=x\left(x^2-3x+1\right)-\left(x^2-3x+1\right)=\left(x-1\right)\left(x^2-3x+1\right)\)

c) \(x^3-3x^2-3x+1=x^3-4x^2+x+x^2-4x+1\)

\(=x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)=\left(x+1\right)\left(x^2-4x+1\right)\)

d) câu này đề hình như sai rồi đó bn