TÌM X biết 1/3-2/3(X+1)=0
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3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
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1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
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a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
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\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\)
\(2x^2-x+\frac{1}{3}-\frac{2}{3}x=0\)
\(2x^2-\frac{5}{3}x+\frac{1}{3}=0\)
\(6x^2-5x+1=0\)
\(6x^2-3x-2x+1\)
\(3x\left(2x-1\right)-\left(2x-1\right)=0\)
\(\left(3x-1\right)\left(2x-1\right)=0\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)
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\(\text{1 , ( x - 3 ) . ( 4 - x ) = 0}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\4-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\in Z\\x=4\in Z\end{cases}}\)
vậy______
\(2,\left(x-5\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\in Z\\x\in\varnothing\end{cases}}\)
vậy x = 5
3, ( x + 1 ) + ( x + 2 ) + (x + 3 ) + ... +( x + 99 ) = 0
(x+x+x+....+x)+(1+2+3+.....+99) = 0
(x.99) + 5050 = 0
x.99 = 0-5050
x.99 = -5050
x = -5050 : 99
x = \(\frac{-5050}{99}\notin Z\Rightarrow x\in\varnothing\)
vậy_____
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Ta có : \(\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}>0\)
- Đặt \(f\left(x\right)=\dfrac{\left(x-3\right)\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x-4\right)}\)
- Lập bảng xét dấu :
- Từ bảng xét dấu : - Để f(x) > 0
\(\Leftrightarrow\left[{}\begin{matrix}-3< x< -2\\-1< x< 3\\x>4\end{matrix}\right.\)
Vậy ...
1/3-2/3(x+1)=0
2/3(x+1)=1/3-0
2/3(x+1)=1/3
x+1=1/3:2/3
x+1=1/2
x=1/2-1
x=-1/2
\(\frac{2}{3}\left(x+1\right)=\frac{1}{3}-0\)
\(\frac{2}{3}\left(x+1\right)=\frac{1}{3}\)
\(\left(x+1\right)=\frac{1}{3}:\frac{2}{3}\)
\(x+1=\frac{1}{2}\)
\(x=\frac{1}{2}-1\)
\(x=\frac{-1}{2}\)