\(\dfrac{1}{2}\) . \(x\) + \(\dfrac{3}{5}\) . (\(x\) - 2) =3
3 - (\(\dfrac{1}{6}\) - \(x\)) . \(\dfrac{2}{3}\) = \(\dfrac{2}{3}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)
1: Ta có: \(\dfrac{x}{3}=\dfrac{y}{6}\)
mà 4x-y=42
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{4x-y}{4\cdot3-6}=\dfrac{42}{12-6}=\dfrac{42}{6}=7\)
=>\(x=7\cdot3=21;y=6\cdot7=42\)
2: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x-2y+3z=33
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2\cdot3+3\cdot5}=\dfrac{33}{2-6+15}=\dfrac{33}{11}=3\)
=>\(x=3\cdot2=6;y=3\cdot3=9;z=3\cdot5=15\)
3: \(\dfrac{x}{y}=\dfrac{6}{5}\)
=>\(\dfrac{x}{6}=\dfrac{y}{5}\)
mà x+y=121
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{x+y}{6+5}=\dfrac{121}{11}=11\)
=>\(x=11\cdot6=66;y=11\cdot5=55\)
\(\dfrac{9}{5}+\dfrac{5}{7}+\dfrac{7}{5}+\dfrac{3}{7}=\left(\dfrac{9}{5}+\dfrac{7}{5}\right)+\left(\dfrac{5}{7}+\dfrac{3}{7}\right)=\dfrac{16}{5}+\dfrac{8}{7}=\dfrac{112}{35}+\dfrac{40}{35}=\dfrac{152}{35}\)
\(\dfrac{1}{2}\times\dfrac{45}{33}\times\dfrac{1}{9}\times\dfrac{11}{6}=\dfrac{1}{2}\times\dfrac{15}{11}\times\dfrac{1}{9}\times\dfrac{11}{6}=\left(\dfrac{1}{2}\times\dfrac{1}{9}\right)\times\left(\dfrac{15}{11}\times\dfrac{11}{6}\right)=\dfrac{1}{18}\times\dfrac{15}{6}=\dfrac{5}{36}\)
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
\(\Leftrightarrow6x-3-5x+10-x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
`@` `\text {Ans}`
`\downarrow`
`1/2*x + 3/5*(x - 2) = 3`
`=> 1/2*x + 3/5*x - 3/5*2 = 3`
`=> 1/2x + 3/5x - 6/5 = 3`
`=> (1/2 + 3/5)x - 6/5 = 3`
`=> 11/10x - 6/5 = 3`
`=> 11/10x = 3 + 6/5`
`=> 11/10x =21/5`
`=> x = 21/5 \div 11/10`
`=> x = 42/11`
Vậy, `x = 42/11`
____
`3 - (1/6 - x)*2/3 = 2/3`
`=> (1/6 - x)*2/3 = 3 - 2/3`
`=> (1/6 - x)*2/3 = 7/3`
`=> 1/6 - x = 7/3 \div 2/3`
`=> 1/6 - x=7/2`
`=> x = 1/6 - 7/2`
`=> x = -10/3`
Vậy, `x = -10/3.`
\(\dfrac{1}{2}\cdot x+\dfrac{3}{5}\left(x-2\right)=3\\ \dfrac{1}{2}\cdot x+\dfrac{3}{5}\cdot x-\dfrac{13}{5}=3\\ \left(\dfrac{1}{2}+\dfrac{3}{5}\right)x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x=\dfrac{28}{5}\\ x=\dfrac{28}{5}:\dfrac{11}{10}\\ x=\dfrac{28}{11}\\ 3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}\\ x=-\dfrac{10}{3}\)