(x-1/2):3=3:(x-1/2) ai làm được mik tick ......
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(X+2)-(x-3)=7x-2(x+1)
2x-1=5x-2
2x-1+1=5x-2+1
2x=5x-1
2x-5x=5x-1-5x
-3x=1
-3x/-3=-1/-3
X=1/3
Vậy x=1/3
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+..+9\right)}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times\left(1+1\right)}{2}+\frac{2\times\left(2+1\right)}{2}+\frac{3\times\left(3+1\right)}{2}...+\frac{9\times\left(9+1\right)}{2}}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times2}{2}+\frac{2\times3}{2}+\frac{3\times4}{2}+...+\frac{9\times10}{2}}{1\times2+2\times3+3\times4+...+9\times10}\)
\(=\frac{\frac{1}{2}\times\left(1\times2+2\times3+3\times4+...+9\times10\right)}{1\times2+2\times3+3\times4+...+9\times10}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)
Dấu \(.\)là dấu nhân
\(y+y.\frac{1}{4}:\frac{2}{7}+y:\frac{2}{9}=255\)
\(\Rightarrow y+y.\frac{1}{4}.\frac{7}{2}+y.\frac{9}{2}=255\)
\(\Rightarrow y+y.\frac{7}{8}+y.\frac{9}{2}=255\)
\(\Rightarrow y.\left(1+\frac{7}{8}+\frac{9}{2}\right)=255\)
\(\Rightarrow y.\left(\frac{8}{8}+\frac{7}{8}+\frac{36}{8}\right)=255\)
\(\Rightarrow y.\frac{51}{8}=255\)
\(\Rightarrow y=255:\frac{51}{8}\)
\(\Rightarrow y=255.\frac{8}{51}\)
\(\Rightarrow y=40\)
Vậy \(y=40\)
~ Ủng hộ nhé
15/2 - ( x * 3 ) = 0,2
x * 3 = 15/2 - 0,2
x * 3 = 37,5
x = 37,5 : 3
x = 12,5
2/3 : x - 6 = 1/2
2/3 : x = 1/2 + 6
2/3 : x = 13/2
x = 2/3 : 13/2
x = 4/39
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
⇔\(\left(x+4\right)\left(x+4\right)=100\)
⇔\(\left(x+4\right)^2=10^2\)
⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
a, 2^(x+1)+2^x=192
=>2^x . 2+2^x=192
=>2^x(2+1)=192
=>2^x=64
=>x=6
Vây x=6
b,2^x . (2^2)^2=(2^3)^2
=>2^x . 4^2=8^2
=>2^x . 16=64
=>2^x=4
=>x=2
Vậy x=2
\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\\ A_{min}=10\Leftrightarrow x-3=0\Leftrightarrow x=3\)
\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi x=3
\(x\left(x-\frac{1}{3}\right)< 0\)
Để \(x\left(x-\frac{1}{3}\right)< 0\)thì x và \(x-\frac{1}{3}\)trái dấu nhau
Thấy \(x>x-\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}x>0\\x-\frac{1}{3}< 0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x< \frac{1}{3}\end{cases}\Leftrightarrow}0< x< \frac{1}{3}}\)
\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
ai biết thì giúp đê :3
\(\frac{x-\frac{1}{2}}{3}=\frac{3}{x-\frac{1}{2}}\) \(\Rightarrow\left(x-\frac{1}{2}\right)^2=3^2\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-3^2=0\)
\(\Rightarrow\left(x-\frac{1}{2}-3\right)\left(x-\frac{1}{2}+3\right)=0\)
\(\Rightarrow\left(x-\frac{7}{2}\right)\left(x+\frac{5}{2}=0\right)\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{2}=0\\x+\frac{5}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{5}{2}\end{cases}}}\)