Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left|x+1\right|+\left|-13\right|=26\)
\(\left|x+1\right|+13=26\)
\(\left|x+1\right|=26-13\)
\(\left|x+1\right|=13\)
\(\Rightarrow\orbr{\begin{cases}x+1=13\\x+1=-13\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-14\end{cases}}\)
vậy ........
b) \(3^{x+2}+3^x=250\)
\(3^x.3^2+3^x=250\)
\(3^x.\left(3^2+1\right)=250\)
\(3^x.10=250\)
\(3^x=25\)
\(\Rightarrow x\in\varnothing\)
Những câu sau tương tự
a, 5x+5x+2=650
<=>5x+5x.52=650
<=>5x.(1+52)=650
<=>5x.26=650
=>5x=650:26
=>5x=25=52
=>x=2
b, 3x-1+5.3x-1=162
<=>3x-1.(5+1)=162
<=>3x-1.6=162
=>3x-1=162:6
=>3x-1=27=33
=>x-1=3
=>x=3+1
=>x=4
Chúc bạn học giỏi nha!!!
K cho mik với nhé Nguyễn Công Đạt
a)\(5^x+5^{x+2}=650\)(=)\(5x.\left(1+25\right)=650\)(=)\(5^x.26=650\)(=)\(5^x=25\)=>x=2
b)\(3^{x-1}+5.3^{x-1}=162\)(=)\(3^{x-1}.\left(1+5\right)=162\)(=)\(3^{x-1}.6=162\)(=)\(3^{x-1}=27\)(=)\(3^{x-1}=3^3\)=>x-1=3(=)x=2
c)\(4^x+4^{x+3}=4160\)(=)\(4^x.\left(1+64\right)=4160\)(=)\(4^x.65=4160\)(=)\(4^x=64\)(=)\(4^x=4^3\)
=>x=3
học tốt
a) \(5^x+5^{x+1}=650\)
\(5^x+5^x.5=650\)
\(5^x.\left(1+5\right)=650\)
\(5^x.6=650\)
\(5^x=\frac{325}{3}\)
kết quả của đề này lẻ quá, bạn xem lại đề câu này nhé
b) \(3^{x-1}+5.3^{x-1}=162\)
\(3^{x-1}.\left(1+5\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
vậy \(x=4\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)
=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
Bài 2. Ta có: (3x - 5)100 \(\ge\)0 \(\forall\)x
(2y + 1)100 \(\ge\)0 \(\forall\)y
=> (3x - 5)100 + (2y + 1)100 \(\ge\)0 \(\forall\)x;y
Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)
Vậy ...
a,\(|x+1|+13=26\)
\(\Rightarrow|x+1|=26-13\)
\(\Rightarrow|x+1|=13\)
TH1: \(x+1=13\)
\(x=13-1\)
\(x=12\)
TH2: \(x+1=-13\)
\(x=\left(-13\right)-1\)
\(x=-14\)
b, \(3^x.\left(3^2+1\right)=250\)
\(3^x.10=250\)
\(3^x=250:10\)
\(3^x=25\)
c,
a) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x.1+5^x.5^2=650\)
\(\Rightarrow5^x.\left(1+5^2\right)=650\)
\(\Rightarrow5^x.\left(1+25\right)=650\)
\(\Rightarrow5^x.26=650\)
\(\Rightarrow5^x=650:26\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\left(TM\right).\)
Vậy \(x=2.\)
b) \(3^{x-1}+5.3^{x-1}=162\)
\(\Rightarrow1.3^{x-1}+5.3^{x-1}=162\)
\(\Rightarrow3^{x-1}.\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}.6=162\)
\(\Rightarrow3^{x-1}=162:6\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1\)
\(\Rightarrow x=4\left(TM\right).\)
Vậy \(x=4.\)
Chúc bạn học tốt!
a) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x\left(1+5^2\right)=650\)
\(\Leftrightarrow5^x=25=5^2\)
\(\Leftrightarrow x=2\)
b) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}\left(1+5\right)=162\)
\(\Leftrightarrow3^{x-1}=27=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)