(1/4x -4)^3+(3/4x+3)^3+(1-x)^3=0
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\(\left|-4x\right|=2\left(x+1\right)\)
\(-4x< 0\Leftrightarrow x>0\)
\(-4x\ge0\Leftrightarrow x\le0\)
Voi: \(x>0\)
\(pt\Leftrightarrow4x=2\left(x+1\right)\Leftrightarrow x=1\left(tm\right)\)
Voi : \(x\le0\)
\(pt\Leftrightarrow-4x=2\left(x+1\right)\Leftrightarrow x=-\frac{1}{3}\left(tm\right)\)
Vay phuong trinh nay co tap nghiem la: \(S=\left\{1;-\frac{1}{3}\right\}\)
Khong chac :<

Ta có: P = AB = \(\frac{3}{4}\cdot\left(\frac{x+3}{x-3}\right)=\frac{3\left(x+3\right)}{4\left(x-3\right)}=\frac{3x+9}{4x-12}\)
=> 4P = \(\frac{4\left(3x+9\right)}{4x-12}=\frac{4\left(3x+9\right)}{4\left(x-3\right)}=\frac{3x+9}{x-3}=\frac{3\left(x-3\right)+18}{x-3}=3+\frac{18}{x-3}\)
Để P \(\in\)Z <=> 4P \(\in\)Z <=> 18 \(⋮\)x - 3
<=> x - 3 \(\in\)Ư(18) = {1; -1; 2; -2; 3; -3; 6; -6; 9; -9; 18; -18}
Lập bảng:
x - 3 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 | 9 | -9 | 18 | -18 |
x | 4 | 2 | 5 | 1 | 6 | 0 | 9 | -3 | 12 | -6 | 21 | -15 |
vậy ....

Thay \(2016=xyz\)vào biểu thức ta được
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+z+1}{xz+z+1}=1\)
Vậy \(A=1\)
Vì \(xyz=2016\)
\(\Rightarrow A=\frac{2016x}{xy+2016x+2016}+\frac{y}{yz+y+2016}+\frac{z}{xz+z+1}\)
\(=\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\)


Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(\Rightarrow A=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=t\)
\(\Rightarrow A=\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)