(7x - 11)3= 100
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`(7x - 11)^2 = 2^5 * 5^2 + 100?`
`=> (7x - 11)^2 = 32*25 + 100`
`=> (7x - 11)^2 = 800 + 100`
`=> (7x - 11)^2 = 900`
`=> (7x - 11)^2 = (+-30)^2`
`=>`\(\left[{}\begin{matrix}7x-11=30\\7x-11=-30\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}7x=41\\7x=-19\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{41}{7}\\x=-\dfrac{19}{7}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{\dfrac{41}{7};-\dfrac{19}{7}\right\}\)
sorry
(7x - 11 ) 3= 32 + 100 . 100 + 1
(7x - 11 ) 3 = 1000
(7x - 11 ) 3 = 103
=> 7x + 11 = 10
=> 7x = 10 + 11 = 21
=> x = 21 : 1 = 3
Đầu bài sai.
Sửa :
\(\left(7x-11\right)^3=10000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
=>7x-11=\(\sqrt[3]{100}\)
=>\(7x=\sqrt[3]{100}+11\)
=>\(x=\dfrac{\sqrt[3]{100}+11}{7}\)