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(x + 1)(x + 2)(x + 3) = x3 - 1
=> x3 + 6x2 + 11x + 6 - x3 + 1 = 0
=> 6x2 + 11x + 7 = 0
Vì 6x2 + 11x + 7 > 0 => vô nghiệm
Vậy \(x\in\phi\)
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\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2\left(x+3\right)}{2x\left(x+3\right)}+\frac{2x}{2x\left(x+3\right)}=\frac{x\left(x+3\right)}{2x\left(x+3\right)}\)
\(\Leftrightarrow2x+6+2x=x^2+3x\)
\(\Leftrightarrow x=3\)
\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{x+x+3}=\frac{1}{2}\)
x+x+3=2
2x=-1
x=-1/2
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Đặt x+1=a; x-2=b
Phương trình trở thành:
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;2;\dfrac{1}{2}\right\}\)
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\(\orbr{\begin{cases}x>3\\x\le0\end{cases}}\)Thì mới thỏa mãn yêu cầu bài
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(x+1)(x+2)(x+3)=x3-1
<=>x.(x+2)(x+3)+(x+2)(x+3)=x3-1
<=>(x2+2x)(x+3)+x.(x+3)+2.(x+3)=x3-1
<=>x2.(x+3)+2x.(x+3)+x2+3x+2x+6=x3-1
<=>x3+3x2+2x2+6x+x2+3x+2x+6=x3-1
<=>x3-x3+3x2+2x2+x2+6x+3x+2x+6+1=0
<=>6x2+17x+7=0
<=>6x2+3x+14x+7=0
<=>3x.(2x+1)+7.(2x+1)=0
<=>(2x+1)(3x+7)=0
<=>2x+1=0 hoặc 3x+7=0
<=>x=-1/2 hoặc x=-7/3
Vậy S={-1/2;-7/3}
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b. `|x + 1| + |2x - 3| = |3x - 2|`
Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)
Vậy phương trình có vô số nghiệm.
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\(\left(x+1\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-1;3\right\}\)
PT ∈ {-1; 3}