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Có: x2+x+1\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) với mọi x
=>x3+x2+x+1>x3
=>y3>x3 (1)
Lại có (x+2)3-(x3+x2+x+1)
=x3+8+6x2+12x-x3-x2-x-1=5x2+11x+7=\(5\left(x^2+\frac{11}{5}x+\frac{7}{5}\right)=5\left(x^2+2.x.\frac{11}{10}+\frac{121}{100}+\frac{19}{100}\right)=5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}\ge\frac{19}{20}>0\) với mọi x
=>(x+2)3 \(\ge\) x3+x2+x+1 (2)
Từ (1),(2)
=>x3<y3<(x+2)3
=>y3=(x+1)3 => x3+x2+x+1=(x+1)3
=>x2(x+1)+(x+1)-(x+1)3=0
=>(x2+1)(x+1)-(x+1)3=0
=>(x+1)x=0=>x=0 hoặc x=-1
+x=0 thì y=1
+x=-1 thì y=0
Vậy (x;y)=...............
(x2+x+1)(x2+x+2)=12
=>x=-2 hoặc 1
x(x+1)(x2+x+1)=42
=>x=-3 hoặc 2
(x2+x+1)2= 3(x4+x2+1)
<=>3(x4+x2+1)=3(x2-x+1)(x2+x+1)
=>(x2+x+1)2=3(x2-x+1)(x2+x+1)
=>x=1
a) \(x^4+x^3+x+1\)
\(\left(x^4+x^3\right)+\left(x+1\right)\)
\(x^3\left(x+1\right)\)+(x+1)
(x+1)(\(x^3+1\))
e)\(ax^2+ay-bx^2-by\)
\(\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(\left(x^2+y\right)\left(a-b\right)\)
\(\left(x+1\right)^3+\left(x-1\right)^3=\left(x-1\right)\left(x+1\right)+4\)
\(\Rightarrow\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x+1\right)\left(x-1\right)-4=0\)
\(\Rightarrow2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)-x^2+1-4=0\)
\(\Rightarrow2x\left(x^2+3\right)-x^2+1-4=0\)
\(\Rightarrow2x^3+6x-x^2-3=0\)
\(\Rightarrow\left(2x^3+6x\right)-\left(x^2+3\right)=0\)
\(\Rightarrow2x\left(x^2+3\right)-\left(x^2+3\right)=0\)
\(\Rightarrow\left(x^2+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=-3\left(L\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)