(x2-1)3+(x2+2)3+(2x-1)3+(3x+3)(2x-1)(1-x)(x2+2)=0
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Lời giải:
PT $\Leftrightarrow (x^2-1)^3+(x^2+2)^3+(2x-1)^3-3(x^2-1)(x^2+2)(2x-1)=0$
Đặt $x^2-1=a; x^2+2=b; 2x-1=c$ thì pt trở thành:
$a^3+b^3+c^3-3abc=0$
$\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0$
$\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$
$\Rightarrow a+b+c=0$ hoặc $a^2+b^2+c^2-ab-bc-ac=0$
Nếu $a+b+c=0$
$\Leftrightarrow x^2-1+x^2+2+2x-1=0$
$\Leftrightarrow 2x^2+2x=0$
$\Rightarrow x=0$ hoặc $x=-1$
Nếu $a^2+b^2+c^2-ab-bc-ac=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$ (dễ CM)
$\Leftrightarrow a=b=c$
$\Leftrightarrow x^2-1=x^2+2=2x-1$ (vô lý)
Vậy..........
Akai Haruma Chị ơi chỗ
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) từ chỗ trên chị tách làm sao ra được vế beeb phải vậy ạ
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a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3+\left(3x+3\right)\left(2x-1\right)\)\(.\left(1-x\right)\left(x^2+2\right)=0\).
\(\Leftrightarrow\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3-3\left(x+1\right)\left(2x-1\right)\)\(.\left(x-1\right)\left(x^2+2\right)=0\).
\(\Leftrightarrow\left(x^2-1\right)^3+\left(x^2+2\right)^3+\left(2x-1\right)^3-3\left(x^2-1\right)\left(2x-1\right)\)\(.\left(x^2+2\right)=0\).
Đặt \(x^2-1=a\), \(x^2+2=b\), \(2x-1=c\). Phương trình trở thành:
\(a^3+b^3+c^3-3abc=0\).
\(\Leftrightarrow\left[a^3+b^3+3ab\left(a+b\right)\right]-3ab\left(a+b\right)+c^3-3abc=0\).
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\).
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\).
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\).
Xét phương trình \(\left(1\right)\).
\(\left(1\right)\Leftrightarrow x^2-1+x^2+2+2x-1=0\)
\(\Leftrightarrow2x^2+2x=0\).
\(\Leftrightarrow2x\left(x+1\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\).
Xét phương trình \(\left(2\right)\)
\(\left(2\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0.2\).
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\).
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\).
\(\Leftrightarrow\left(x^2-1-x^2-2\right)^2+\left(x^2+2-2x+1\right)^2+\left(2x-1-x^2+1\right)^2=0\).
\(\Leftrightarrow\left(-3\right)^2+\left[\left(x-1\right)^2+2\right]^2+\left(-x^2+2x\right)^2=0\).
\(\Leftrightarrow\left[\left(x-1\right)^2+2\right]^2+\left(x^2-2x\right)^2+9=0\).
Ta có:
\(\left(x-1\right)^2\ge0\forall x\).
\(\Rightarrow\left(x-1\right)^2+2\ge2\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2\ge4\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9\ge13\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9>0\forall x\).
\(\left(x^2-2x\right)^2\ge0\forall x\).
\(\Rightarrow\left[\left(x-1\right)^2+2\right]^2+9+\left(x^2-2x\right)^2>0\forall x\).
Do đó phương trình \(\left(2\right)\)vô nghiệm.
Vậy phương trình có tập nghiệm: \(S=\left\{0;-1\right\}\).
\(\)