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\(2x+3=15\\ \Rightarrow2x=15-3=12\\ \Rightarrow x=\dfrac{12}{2}=6\\ ---\\ 28-3x=13\\ \Rightarrow3x=28-13=15\\ \Rightarrow x=\dfrac{15}{3}=5\\ ---\\ \left(x-1954\right).5=50\\ \Rightarrow x-1954=50:5=10\\ \Rightarrow x=10+1954=1964\\ ---\\ x-152:2=46\\ \Rightarrow x-76=46\\ \Rightarrow x=46+76=122\)
a, 2\(x\) + 3 = 15
2\(x\) = 12
\(x\) = 6
b, 28 - 3\(x\) = 13
3\(x\) = 28 - 13
3\(x\) = 15
\(x\) = 5
`a,`
`x*3+7=16`
`=>3x = 16 - 7`
`=> 3x = 9`
`=> x = 9 \div 3`
`=> x = 3`
Vậy, `x = 3`
`b,`
` x- 152 \div 2 = 46?`
`=> x - 76 = 46`
`=> x = 46 + 76`
`=> x = 122`
Vậy, `x = 122.`
`c,`
` 74-2*(x+3)=34`
`=> 2(x + 3) = 74 - 34`
`=> 2(x+2) = 40`
`=> x + 2 = 40 \div 2`
`=> x + 2 = 20`
`=> x = 20 - 2`
`=> x = 18`
Vậy, `x = 18.`
( x - 5 ) * 3 - 140 : 35 = 152
( x - 5 ) * 3 - 4 = 152
( x - 5 ) * 3 = 152 + 4
( x - 5 ) * 3 = 156
x - 5 = 156 : 3
x - 5 = 52
x = 52 + 5
( x + 1 ) + ( x + 5 ) + ... + ( x + 29 ) = 136
Dãy số trên gồm:( 29 - 1 ) : 4 + 1 = 8 ( số hạng )
Ta có : ( x + 29 + x + 1 ) x 8 : 2 = 136
( x * 2 + 30 ) x 4 = 136
x * 2 + 30 = 136 : 4
x * 2 + 30 = 34
x * 2 = 34 - 30
x * 2 = 4
x = 4 : 2
x = 2
Đây là bài toán lớp 4 chứ không phải là lớp 6
( x - 5 ) * 3 - 140 : 35 = 152
( x - 5 ) * 3 - 4 = 152
( x - 5 ) * 3 = 152 + 4
( x - 5 ) * 3 = 156
x - 5 = 156 : 3
x - 5 = 52
x = 52 + 5
( x + 1 ) + ( x + 5 ) + ... + ( x + 29 ) = 136
Dãy số trên gồm:( 29 - 1 ) : 4 + 1 = 8 ( số hạng )
Ta có : ( x + 29 + x + 1 ) x 8 : 2 = 136
( x * 2 + 30 ) x 4 = 136
x * 2 + 30 = 136 : 4
x * 2 + 30 = 34
x * 2 = 34 - 30
x * 2 = 4
x = 4 : 2
x = 2
a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
0,5 x + 0,6 ( x - 2 ) = 3
0,5 x + 0.6 x - 1,2 = 3
1,1 x = 4,2
x = \(\frac{42}{11}\)
Kết luận:
b) \(\frac{1}{3}x-0,5x=0,75\)
\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(-\frac{1}{6}x=\frac{3}{4}\)
\(x=-\frac{9}{2}\)
Kết luận:
c) \(\frac{3}{-2}x-0,5x=75\%\)
-1,5x - 0,5x = 0,75
-2x = 0,75
x = -0,375
Kết luận:
d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
-0,4 x + 0,25 = 0,75 - 0,75 x
-0,4 x + 0,75 x = 0,75 - 0,25
0,35 x = 0,5
x = \(\frac{10}{7}\)
Kết luận:
\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)
\(\frac{11}{10}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)
\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(\frac{-1}{6}x=\frac{3}{4}\)
\(x=\frac{-9}{2}\)
\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)
\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)
\(-2x=\frac{3}{4}\)
\(x=\frac{-3}{8}\)
\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)
\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)
\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)
\(\frac{7}{20}x=\frac{1}{2}\)
\(x=\frac{10}{7}\)
n: (-8)*x+17=-23
=>\(x\cdot\left(-8\right)=-23-17=-40\)
=>\(x\cdot8=40\)
=>\(x=\dfrac{40}{8}=5\)
o: \(\left(-15\right)\cdot x=10\left(-4\right)-5\)
=>\(x\cdot\left(-15\right)=-40-5=-45\)
=>\(x\cdot15=45\)
=>\(x=\dfrac{45}{15}=3\)
p: \(\left(-3\right)\cdot x-4=2\cdot\left(-7\right)+4\)
=>\(\left(-3\right)\cdot x-4=-14+4=-10\)
=>\(x\left(-3\right)=-10+4=-6\)
=>\(3x=6\)
=>\(x=\dfrac{6}{3}=2\)
q: x+x+x+91=-2
=>\(3x+91=-2\)
=>\(3x=-2-91=-93\)
=>\(x=-\dfrac{93}{3}=-31\)
r: \(-152-\left(3x+1\right)=\left(-2\right)\cdot\left(-27\right)\)
=>\(-152-3x-1=54\)
=>\(-153-3x=54\)
=>\(3x=-153-54=-207\)
=>\(x=-\dfrac{207}{3}=-69\)
\(x:0,5+x\) . \(\dfrac{1}{5}=1\dfrac{3}{5}\)
\(x :\)\(\dfrac{1}{2}+x.\dfrac{1}{5}=\dfrac{8}{5}\)
\(x . 2+x.\)\(\dfrac{1}{5}=\dfrac{8}{5}\)
\(x\)\((\)\(2+\)\(\dfrac{1}{5}\)\()\) \(=\)\(\dfrac{8}{5}\)
\(x.\)\(\dfrac{11}{5}=\dfrac{8}{5}\)
\(x=\)\(\dfrac{8}{5}:\dfrac{11}{5}\)
\(x=\)\(\dfrac{8}{11}\)
vậy \(x=\)\(\dfrac{8}{11}\)