Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
b1
ta có : n+4 = (n+1)+3
=>n+1+3 chia hết cho n+1
vì n+1 chia hết cho n+1
=>3 chia hết cho n+1
=> n+1 chia hết cho 3
=> n+1 thuộc Ư 3 =[1;3]
=> n+1=1 n+1=3
n =1-1 n =3-1
n =0 n =2
vậy n thuộc [0;2]
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
a. 3x+3x+1+3x+2=1053
=> 3x.(1+3+32)=1053
=> 3x.(1+3+9)=1053
=> 3x.13=1053
=> 3x=1053:13
=> 3x=81
=> 3x=34
=> x=4
b. 5x.519=520.511
=> 519+x=520+11
=> 19+x=20+11
=> 19+x=31
=> x=31-19
=> x=12
\(3^x+3^{x+1}+3^{x+2}=1053\)
\(3^x\left(1+3+3^2\right)=1053\)
\(3^x\cdot13=1053\)
\(3^x=1053:13=81\)
\(3^x=3^4\)
=> x = 4
+) \(x^3=x^2\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
+) \((7x-11)^3=2^5.5^2+200\)
\((7x-11)^3=2^3.2^2.5^2+2^3.5^2\)
\((7x-11)^3=2^3.5^2.(2^2+1)\)
\((7x-11)^3=2^3.5^2.5\)
\((7x-11)^3=2^3.5^3\)
\((7x-11)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(7x=21\)
\(x=3\)
+) \(3+2^{x-1}=24-[4^2-(2^2-1)]\)
\(3+2^{x-1}=11\)
\(2^{x-1}=8\)
\(2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(x=4\)
\(\text{a) ( x + 1 ) + ( x + 3 ) + ..... + ( x + 99 ) = 0}\)
\(\Rightarrow\left(x+x+x+.....+x\right)+\left(1+3+5+....+99\right)=0\)
\(\text{Ta có :}\)
\(1+3+5+...+99=\frac{\left(99-1\right):2+1.\left(99+1\right)}{2}=2500\)
\(\Rightarrow50x+2500=0\)
\(\Rightarrow50x=-2500\)
\(\Rightarrow x=-50\)