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a, \(3x+12=2\)
\(\Rightarrow3x=-10\)
\(\Rightarrow x=\frac{-10}{3}\)
=>\(\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(2x\right)}{18}+\frac{2\left(-5x+3\right)}{18}=\frac{1}{2}\)
=>\(\frac{6x-6}{18}+\frac{27x-45}{18}+\frac{4x}{18}+\frac{-10x+6}{18}=\frac{1}{2}\)
=>\(\frac{\left(6x-6\right)+\left(27x-45\right)+4x+\left(-10x+6\right)}{18}=\frac{1}{2}\)
=>\(\frac{27x-45}{18}=\frac{1}{2}\)
=>54x-90=18
54x =108
=>x=2
\(\left(X+5\right)-\left(X-9\right)=X+2\)\(2\)
\(=>X+5-X+9=X+2\)
\(=>\left(X-X\right)+\left(5+9\right)=X+2\)
\(=>0+14=X+2\)
\(=>14=X+2\)
\(=>X=12\)
(x+5)-(x-9)=x+2
x+5-x+9=x+2
x-x+14=x+2
12=x(cũng bớt mới về đi 2 đơn vị)
hoặc x=14-2suy ra x=12
bài 1
\(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(\Rightarrow\left(3x-y\right).4=\left(x+y\right)3\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
Vậy \(\frac{x}{y}=\frac{7}{9}\)
Bài 1 : Ta có:
\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)
= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)
= \(\frac{7}{9}\)
Bài 2 :
\(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)
=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)
=> 50x = 10
=> x = 10 : 50
=> x = 1/5
Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
a) \(3x-\frac{3}{2}-5x+\frac{10}{3}=1\)
\(3x-5x=1+\frac{3}{2}-\frac{10}{3}\)
\(-2x=-\frac{5}{6}\)
\(x=\frac{5}{12}\)
b) \(\left|x-1\right|=5-x\)
Th1:
\(x-1=5-x\)
\(x+x=5+1\)
\(2x=6\)
\(x=3\)
Th2:
\(-\left(x-1\right)=5-x\)
\(x+1=5-x\)
\(x+x=5-1\)
\(2x=4\)
\(x=2\)
Vậy \(x=3\)và \(x=2\)
a) 3x - 5x = 1 + 3/2 - 10/3
-2x = -5/6
x = -5/6 : ( - 2 )
x = 5/12
b) |x-1|= 5-x
Nếu x \(\ge\)1 \(\Rightarrow\)x - 1 \(\ge\)0 \(\Rightarrow\)x - 1 = 5 - x.
2x = 6
x = 3.
Nếu x < 1 \(\Rightarrow\)x - 1 < 0 \(\Rightarrow\)Ix-1I = 1 - x
\(\Rightarrow\)1 - x = 5 - x \(\Rightarrow\)vô lý.
Vậy x = 3
\(3x-\frac{3}{2}-5x+\frac{10}{3}=1\)
\(3x-5x=1+\frac{3}{2}-\frac{10}{3}\)
\(-2x=-\frac{5}{6}\)
\(x=\frac{5}{12}\)
=.= hok tốt !!
<=>\(-\frac{21}{6}-2x=1\)
<=>\(-2x=1+\frac{21}{6}\)
<=>\(x=\frac{27}{6}:-2\)
<=>\(x=-\frac{27}{12}\)