TÌm `x` :
`3(x-1/2)-2(x+1/3)=0`
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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}
1.
$(3^2-2^3)x+3^2.2^2=4^2.3$
$\Leftrightarrow x+36=48$
$\Leftrightarrow x=48-36=12$
2.
$x^5-x^3=0$
$\Leftrightarrow x^3(x^2-1)=0$
$\Leftrightarrow x^3(x-1)(x+1)=0$
$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.
$(x-1)^2+(-3)^2=5^2(-1)^{100}$
$\Leftrightarrow (x-1)^2+9=25$
$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$
$\Rightarrow x-1=4$ hoặc $x-1=-4$
$\Leftrightarrow x=5$ hoặc $x=-3$
4.
$(2x-1)^2-(2x-1)=0$
$\Leftrightarrow (2x-1)(2x-1-1)=0$
$\Leftrightarrow (2x-1)(2x-2)=0$
$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$
$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$
$\Lef
`@` `\text {Ans}`
`\downarrow`
\((3^2-2^3)x+3^2.2^2=4^2.3\)
`=> x + (3*2)^2 = 48`
`=> x+6^2 = 48`
`=> x + 36 = 48`
`=> x = 48 - 36`
`=> x=12`
Vậy, `x=12`
\(x^5-x^3=0\)
`=> x^3(x^2 - 1)=0`
`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy, `x \in {0; +- 1 }`
\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)
`=> (x-1)^2 + 9 = 25*1`
`=> (x-1)^2 + 9 = 25`
`=> (x-1)^2 = 25 - 9`
`=> (x-1)^2 = 16`
`=> (x-1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy, `x \in {5; -3}`
\((2x-1)^2-(2x-1)=0\)
`=> (2x-1)(2x-1) - (2x-1)=0`
`=> (2x-1)(2x-1-1)=0`
`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow x-1=0\) hoặc \(x+1=0\)
\(\Leftrightarrow x=1\) hoặc \(x=-1\)
c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
(do \(x^2+1\ge1>0\))
1) \(x\left(x-1\right)-2\left(1-x\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x\left(x+1\right)=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
2) \(\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
3) \(\left(x-3\right)^3+\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)^3+3=x\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-2\right)=0\)
4) \(x^3=x^5\)
\(\hept{\begin{cases}x=-1\\x=1\\x=0\end{cases}}\)
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
quào chữ đẹp thế a :v