(8.x-16) . (x-5) =0
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![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(\left(8\times x-16\right)\times\left(x-5\right)=0\)
=> \(8\times x-16=0\) hoặc \(x-5=0\)
TH1:
\(8\times x-16=0\)
\(8\times x=16\)
\(x=16\div8\)
\(x=2\)
TH2:
\(x-5=0\)
\(x=5\)
Vậy \(x\in\left\{2;5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(8 - 16)(\(x\) - 5) = 0
\(x\) - 5 = 0
\(x\) = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
0,81x8,4+2,6x0,81
= 0,81x(8,4+2,6)
=0,81x11
=8,91
16,5x47,8+47,8x3,5
=47,8x(16,5+3,5)
=47,8x20
=956
nho h cho mk nhe
a, 0,81 x 8,4 + 2,6 x 0,81
= 0,81 x ( 8,4 + 2,6 )
= 0,81 x 11
= 8,91
b, 16,5 x 47,8 + 47,8 x 3,5
= 47,8 x ( 16,5 + 3,5 )
= 47,8 x 20
= 956
![](https://rs.olm.vn/images/avt/0.png?1311)
1)=>3(x-5)(2x+9)+3(x-5)=0=>(x-5)(6x+30)
=>x-5=0=>x=5
6x+30=0=>x=-5
2)=>x^2-16=0=>x=+-4
12-4x=0=>x=3
3)=>9-x^2=0=>x=+-3
4x-8=0=>x=2
4)=>8-x^3=0=>x=3
5^x-125=0=>x=2
5)=>2^x.2^x=8=>2^2x=8=>2x=3=>x=1,5
![](https://rs.olm.vn/images/avt/0.png?1311)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}26⋮x\\x\ge13\end{matrix}\right.\Rightarrow x\in\left\{13;26\right\}\)
\(\left\{{}\begin{matrix}16⋮x\\x< 8\end{matrix}\right.\Rightarrow x\in\left\{1;2;4\right\}\)
\(\left\{{}\begin{matrix}18⋮x\\0< x< 40\end{matrix}\right.\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)
\(\left\{{}\begin{matrix}x⋮15\\30< x< 40\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮12\\22\le5x\le50\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮4\\16\le x\le36\end{matrix}\right.\Rightarrow x\in\left\{16;20;24;28;32;36\right\}\)
Ta có :
(8.x-16) . (x-5) =0
\(\Rightarrow\orbr{\begin{cases}8x-16=0\\x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}8x=16\\x=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
ta có :
\(\left(8x-16\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}8x-16=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
vậy hoặc x=2 hoặc x=5