Question

Tìm x

a)  (245 - x) + 7² = 149

b) x . (x + 5) = 0

c) 2x . (x + 3) = 0

d) 71. 2 - 6. (2x + 5) = 10⁵ : 10³

e) 5 + 3^x+1 = 86

Answer 1

`a, (245 - x) + 7^2 = 149`

`=> (245 - x) + 49 = 149`

`=> 245 - x = 149 - 49`

`=> 245 - x = 100`

`=> x = 245 - 100`

`=> x = 145`

Vậy `x = 145`

\(b,x\cdot\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy `x in {0;-5}`

^{`d,71 * 2 - 6 * (2x + 5) = 10^5 : 10^3`}

^{`=> 142 - 6*(2x + 5) = 100`}

^{`=> 6*(2x + 5) = 142 - 100`}

^{`=> 6*(2x + 5) = 42`}

^{`=> 2x + 5 = 42 : 6`}

^{`=> 2x + 5 = 7`}

^{`=> 2x = 7 - 5`}

^{`=> 2x = 2`}

^{`=> x = 2 :2`}

^{`=> x = 1`}

^{Vậy `x = 1`}

 

 

Answer 2

`e, 5 + 3^{x+1} = 86`

`=> 3^{x+1} = 86 - 5`

`=> 3^{x+1} = 81`

`=> 3^{x+1} =3^4`

`=> x + 1 = 4`

`=> x = 4 - 1`

`=> x = 3`

vậy...

Answer 3

\(c,2x.\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{0;-3\right\}\)