Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\left(1-x\right)^3=216\)
\(\Rightarrow1-x=6\)
\(\Rightarrow x=-5\)
b)\(3^{x+1}-3^x=162\)
\(\Rightarrow3^x\left(3-1\right)=162\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
c)\(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d) \(\left(2x+1\right)^3=-64\)
=> \(\left(2x+1\right)^3=\left(-4\right)^3\)
=> \(2x+1=-4\)
=> \(2x=\left(-4\right)-1\)
=> \(2x=-5\)
=> \(x=\left(-5\right):2\)
=> \(x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}.\)
Mình chỉ làm câu d) thôi nhé.
Chúc bạn học tốt!
Gửi bạn Velvet Red ! Hơi bận nên chỉ làm hai phần a,b cho bạn thôi nhé !!
a) \(3^{x+2}-3^{x+1}=162\)
\(\Leftrightarrow3^x.3^2-3^x.3=162\)
\(\Leftrightarrow3^x.\left(3^2-3\right)=162\)
\(\Leftrightarrow3^x=162:6\)
\(\Leftrightarrow3^x=27=3^3\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
b) \(5^{x+2}-2.5^{x+1}=375\)
\(\Leftrightarrow5^x.5^2-2.5^x.5=375\)
\(\Leftrightarrow5^x.\left(5^2-2.5\right)=375\)
\(\Leftrightarrow5^x=375:15\)
\(\Leftrightarrow5^x=25=5^2\)
\(\Leftrightarrow x=2\)
Vậy : \(x=2\)
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
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, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
a,-12(x-5)+7(3-x)=20
-12x+60+21-7x=20
-19x=-61
x=\(\frac{61}{19}\)
b,30(x+1)-3(x-5)-15x=25
30x+30+15-3x-15x=25
12x=-20
x=\(-\frac{20}{12}\)
a) \(3^{x+1}-3^x=162\)
\(\Leftrightarrow3^x.\left(3-1\right)=162\)
\(\Leftrightarrow3^x.2=162\)
\(\Leftrightarrow3^x=162:2=81\)
\(\Leftrightarrow3^x=3^4\)
\(\Leftrightarrow x=4\)
b) \(\left(1-x\right)^3=216\)
\(\Leftrightarrow\left(1-x\right)^3=6^3\)
\(\Leftrightarrow1-x=6\)
\(\Leftrightarrow x=1-6\)
\(\Leftrightarrow x=-5\)
c) \(5^{x+1}-2.5^x=375\)
\(\Leftrightarrow5^x.\left(5-2\right)=375\)
\(\Leftrightarrow5^x.3=375\)
\(\Leftrightarrow5^x=375:3=125\)
\(\Leftrightarrow5^x=5^3\)
\(\Leftrightarrow x=3\)
thanh bn nha