TÌM X:
3.x+8.x=121
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a. ( 3x - 7 ) 5 = 32
=> 3x - 7 = 32 : 5
=> 3x - 7 = 32/5
=> 3x = 32/5 + 7
=> 3x = 67/5
=> x = 67/5 : 3
=> x = 67/5 . 1/3
=> x = 67/15
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a) ( 3x - 7 ) . 5 = 32
( 3x - 7 ) = 32 : 5
( 3x - 7 ) = 6,4
3x = 6,4 + 7
3x = 13.4
x = 13,4 : 3
x = 4,4666 ..
c, \(6\cdot x+3\cdot x-4=105\)
\(6\cdot x+3\cdot x=105+4\)
\(6\cdot x+3\cdot x=109\)
\(\left(6+3\right)\cdot x=109\)
\(9\cdot x=109\)
\(x=109:9\)
\(x=\frac{109}{9}\)
d, \(200-\left(8\cdot x+7\right)=121\)
\(8\cdot x+7=200-121\)
\(8\cdot x+7=79\)
\(8\cdot x=79-7\)
\(8\cdot x=72\)
\(x=72:8\)
\(x=9\)
\(c,6.x+3.x-4=105\)
\(\left(6+3\right).x=105+4\)
\(9x=109\)
\(x=\frac{109}{9}\)
\(d,200-\left(8.x+7\right)=121\)
\(8.x+7=200-121\)
\(8.x+7=79\)
\(8.x=72\)
\(x=9\)
`a, (4x+5) : 3 -121:11=14`
`=> (4x+5) : 3 - 11=14`
`=> (4x+5) : 3 =14+11`
`=> (4x+5) : 3 = 25`
`=> 4x+5=25xx3`
`=>4x+5= 75`
`=> 4x=75-5`
`=>4x=70`
`=>x= 70/4`
`=>x= 35/2`
`b, 2+4+6+...+x=2450`
Số lượng của dãy là :
`(x-2)/2 + 1= (x-2)/2 +2/2= x/2`
Tổng số lượng là :
\(\dfrac{\left(x+2\right)\cdot\dfrac{x}{2}}{2}=\dfrac{\dfrac{x^2}{2}+\dfrac{2x}{2}}{2}=\dfrac{x\left(x+2\right)}{4}\)
\(\dfrac{x\left(x+2\right)}{4}=2450\)
\(\Rightarrow x\left(x+2\right)=2450\cdot4\\ \Rightarrow x\left(x+2\right)=9800\)
`=>x=98`
`c,` `1` nhân `32` sao?
`d, 2x+3x=1505`
`=> (2+3)x=1505`
`=> 5x=1505`
`=> x= 1505:5`
`=>x= 301`
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
\(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x-\frac{1}{2}=0\)
<=> \(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
<=> \(\hept{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(\left(2x+3\right)^2=\frac{9}{121}\)
<=-> \(\hept{\begin{cases}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{cases}}\)
<=> \(\hept{\begin{cases}2x=\frac{-30}{11}\\2x=\frac{-36}{11}\end{cases}}\)
\(2x^{10}=25x^8\)
<=> \(2x^{10}-25x^8=0\)
<=> \(x^8.\left(2x^2-25\right)=0\)
<=> \(\hept{\begin{cases}x^8=0\\2x^2-25=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x^2=\frac{25}{2}\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{25}{2}}\\x=-\sqrt{\frac{25}{2}}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{cases}}\)
`(3*x+2)^2=121`
\(=>\left[{}\begin{matrix}3x+2=11\\3x+2=-11\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=11-2\\3x=-11-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=9\\3x=-13\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{13}{3}\left(loại\right)\end{matrix}\right.\)
`#3107.101107`
a)
\(x+x+\dfrac{1}{2}\times\dfrac{2}{5}+x+\dfrac{8}{10}=121\\3x+\dfrac{1}{5}+\dfrac{4}{5}=121\\ 3x+1=121\\ 3x=121-1\\ 3x=120\\ x=40 \)
Vậy, `x = 40`
b)
\(\dfrac{12+x}{42}=\dfrac{5}{6}\\ \dfrac{12+x}{42}=\dfrac{35}{42}\\ \dfrac{12+x}{42}-\dfrac{35}{42}=0\\ \dfrac{12+x-35}{42}=0\\ \dfrac{x-\left(35-12\right)}{42}=0\\ \dfrac{x-23}{42}=0\\ x-23=0\\ x=23\)
Vậy,` x = 23.`
a: \(x+x+\dfrac{1}{2}\cdot\dfrac{2}{5}+x+\dfrac{8}{10}=121\)
=>\(3x+\dfrac{1}{5}+\dfrac{4}{5}=121\)
=>3x+1=121
=>3x=120
=>x=40
b: \(\dfrac{x+12}{42}=\dfrac{5}{6}\)
=>\(x+12=42\cdot\dfrac{5}{6}=35\)
=>x=35-12=23
3 . x + 8 . x = 121
3x + 8x = 121
11x = 121
x = 11
\(3.x+8.x=121\)
\(\left(3+8\right).x=121\)
\(11x=121\)
\(x=121:11\)
\(x=11\)