Tìm giá trị nguyên của x để đa thức \(2x^3+11x^2-6x-14⋮x+4\)
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2. Ta có: P = 2x2 + y2 - 4x - 4y + 10
P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4
P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)4 \(\forall\)x;y
=> P luôn dương với mọi biến x;y
3 Ta có:
(2n + 1)(n2 - 3n - 1) - 2n3 + 1
= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1
= -5n2 - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z

a) A = x(y - z) + 2(z - y) = x(y - z) - 2(y - z) = (x - 2)(y - z) = (2 - 2)(1,007 - (-0,006)] = 0
b) B = 2x(y - z) + (z - y)(x + t) = 2x(y - z) - (y - z)(x + t) = (2x - x - t)(y - z) = (x - t)(y - z) = [18,3 - (-31,7)](24,6 - 10,6) = 50.14 = 700
c) C = (x - y)(y + z) + y(y - x) = (x - y)(y + z) - y(x - y) = (x - y)(y + z - y) = (x - y).z = (0,86 - 0,26).1,5 = 0,6.1,5 = 0,9

\(b)4x\left(x-2014\right)-\left(x-2014\right)=0\)
\(\left(4x-1\right)\left(x-2014\right)=0\)
\(\Leftrightarrow TH1:4x-1=0\)
\(4x=1\)
\(x=\frac{1}{4}\)
\(TH2:x-2014=0\)
\(x=2014\)
Vậy \(x\in\left\{\frac{1}{4};2014\right\}\)
\(b,4x\left(x-2014\right)-x+2014=0\)
\(\Leftrightarrow\left(x-2014\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2014\\x=\frac{1}{4}\end{cases}}\)
\(c,\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\left(x+1\right)x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

\(1.=5xy\left(x-2y\right)\)
\(2.=\left(5-y\right)\left(x-y\right)\)
\(3.=y\left(x-z\right)-7\left(x-z\right)=\left(y-7\right)\left(x-z\right)\)
\(5.=2x\left(3y-7z\right)-6y\left(3y-7z\right)=\left(2x-6y\right)\left(3y-7x\right)\)
\(4.=27x^2\left(y-1\right)+9x^3\left(y-1\right)=9x^2\left(3+x\right)\left(y-1\right)\)

\(30xy-16x^2-9y^2\)
\(=-\left(16x^2-24xy+9y^2\right)+6xy\)
\(=-\left(4x-3y\right)^2+6xy\)
\(=-\left[\left(4x-3y\right)^2-6xy\right]\)
\(=-\left(4x-3y-\sqrt{6xy}\right)\left(4x-4y+\sqrt{6xy}\right)\)

Ta có: \(2005\equiv-1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}\equiv-1\left(mod2006\right)\)
Lại có: \(2007=1\left(mod2006\right)\)
\(\Rightarrow2007^{2005}\equiv1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}+2007^{2005}\equiv0\left(mod2006\right)\)
Vậy \(2005^{2007}+2007^{2005}⋮2006\left(đpcm\right)\)