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Ta có: \(\frac{2x-9}{240}=\frac{39}{80}=\frac{117}{240}\)
=>\(2x-9=117\)
\(2x=117+9=126\)
\(x=126:2=63\)
\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\)\(\Rightarrow\)\(2x-9\)=\(\dfrac{240.39}{80}\)=\(117\)
\(2x-9=117\)\(\Rightarrow\)\(2x=117+9=126\)\(\Rightarrow\)\(x=126:2=63\)
\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\) \(\Rightarrow\) \(2x-9=\)\(\dfrac{240.39}{80}\)=117
\(2x-9=117\)\(\Rightarrow\)\(2x=117-9=108\)\(\Rightarrow\)\(x=108:2=54\)
\(\dfrac{2x-9}{240}=\dfrac{117}{240}\)
=> 2x -9=117
2x=117+9
2x=126
x=126/2
x=63
\(\dfrac{2x-9}{240}\)=\(\dfrac{117}{240}\)
►2x-9=117
2x=117+9
2x=126
x=126:2
x=63
Vậy:x=63
Ta có:\(\frac{2x-9}{240}=\frac{39}{40}\Rightarrow\frac{2x-9}{240}=\frac{234}{240}\)
=>2x-9=234
=>2x=234+9
=>2x=243
=>x=243:2
=>x=121,5
Mà x là số nguyên nên không có x thỏa mãn
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
b) \(\dfrac{x}{27}=\dfrac{3}{x}\)
\(\Rightarrow x.x=27.3\)
\(\Rightarrow x^2=81\)
\(\Rightarrow x^2=9^2\)
\(\Rightarrow x=9\)
Vậy x=9
a) x.21=6.7
x.21=42
x=42:21
x = 2
b) y . 20 = -5.28
y.20 = -140
y = (-140) : 20
y = -7
a)=>x*21=7*6
=>x*21=42
=>x=42/21
x=2
b)=>y*20=(-5)*28
=>y*20=-140
=>y=-140/20
y=-7
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Rightarrow\left(2x-9\right).80=240.39\)
\(160x-720=9360\)
\(160x=9360+720\)
\(160x=10080\)
\(x=10080:160\)
\(x=63\)
Vậy x=63
1. Ta có : \(C=\frac{1010}{1008\cdot8-994}=\frac{1010}{\left(1010-2\right)\cdot8-994}=\frac{1010}{1010\cdot8-16-994}\)
\(=\frac{1010}{1010\cdot8-1010}=\frac{1010}{1010\left(8-1\right)}=\frac{1}{7}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot2\cdot3\cdot3+2\cdot4\cdot6\cdot3+3\cdot6\cdot9\cdot3+5\cdot10\cdot15\cdot3}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{3\left(1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15\right)}=\frac{1}{3}\)
2. \(\frac{2x-9}{240}=\frac{39}{80}\)
=> \(\frac{2x-9}{240}=\frac{39\cdot3}{80\cdot3}=\frac{117}{240}\)
=> 2x - 9 = 117
=> 2x = 117 + 9 = 126
=> x = 126 : 2 = 63
nó phải có = bao nhiêu thì mới ra kết quả bạn ạ
\(\dfrac{2x-9}{240}\) \(=\) \(\dfrac{39}{80}\) \(\Leftrightarrow\) \(\left(2x-9\right)\) \(.80\) \(=\) \(240.39\)
\(2x\) \(-\) \(9\) \(=\) ( \(240\) \(.\) \(39\) ) \(:\) \(80\)
\(2x\) \(-\) \(9\) \(=\) \(117\)
\(2x\) \(=\) \(126\)
\(x\) \(=\) \(63\)