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`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2}{3}+\left[\dfrac{4}{5}x-\dfrac{11}{15}\right]=\dfrac{5}{9}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=\dfrac{5}{9}-\dfrac{2}{3}\)
`=>`\(\dfrac{4}{5}x-\dfrac{11}{15}=-\dfrac{1}{9}\)
`=>`\(\dfrac{4}{5}x=-\dfrac{1}{9}+\dfrac{11}{15}\)
`=>`\(\dfrac{4}{5}x=\dfrac{28}{45}\)
`=>`\(x=\dfrac{28}{45}\div\dfrac{ 4}{5}\)
`=>`\(x=\dfrac{7}{9}\)
Vậy, `x = 7/9.`
\(\left(\dfrac{3}{4}-3x\right)\cdot\left(1+4x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{3}{4}-3x=0\\1+4x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}3x=\dfrac{3}{4}\\4x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy, `x \in {1/4; -1/4}.`
a,
Ta có:
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{55}{60}-\frac{60\left(\frac{2}{5}+x\right)}{60}=\frac{40}{60}\)
\(\Leftrightarrow\frac{55}{60}-\frac{24+60x}{60}=\frac{40}{60}\)
=> 55-24-60x=40
<=> 31-60x=40
<=> x=-3/20
Mấy câu còn lại cũng tương tự thế đó
a. 4(2x+7)-3(3x-2)=24
=> 8x+28-9x+6=24
=> -x+34=24
=> -x=-10=> x=10
b. 2(x-1)+3(x-2)=x-4
=> 2x-2+3x-6=x-4
=> 5x-8=x-4
=> 5x-x=-4+8
=> 4x=4=> x=1
c. [124-(20-4x)]:30+7=11
=> (124-20+4x):30=4
=> (104+4x):30=4
=> 104+4x=120
=> 4x=96 => x=24