`1/8+(2x+5)/24=(3x+5)/2+x/6`
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30 tháng 11 2021
a, 2.x + 7 = 15
2x = 8
x = 4
b, 25 – 3.(6 – x) = 22
3.(6-x) = 3
6-x = 1
x = 5
c, [(2x – 11) : 3 + 1].5 = 20
(2x-11) : 3+1 = 4
(2x-11):3 = 3
2x-11 = 1
2x = 12
x = 6
e, 2 . 3x = 10 . 312 + 8 . 274
6x = 3120 + 2192
6x = 5312
x = 5312/6
g, x – 12 = (–8) + (–17)
x - 12 = -25
x = -13
30 tháng 11 2021
Lần sau tách nhỏ nội dung câu hỏi ra nha em, chứ trả lời thế này biếng lắm '^^ Chị làm chỉ mang tính tham khảo kết quả thôi, còn cụ thể thì em tách từng bước một ra he :>
a, \(2\cdot x+7=15\)
\(\Leftrightarrow2\cdot x=8\)
\(\Leftrightarrow x=4\)
Vậy x = 4.
b, \(25-3\cdot\left(6-x\right)=22\)
\(\Leftrightarrow3\cdot\left(6-x\right)=3\)
\(\Leftrightarrow6-x=1\)
\(\Leftrightarrow x=5\)
Vậy x = 5.
c, \(\left[\left(2x-11\right):3+1\right]\cdot5=20\)
\(\Leftrightarrow\left(2x-11\right):3+1=4\)
\(\Leftrightarrow\left(2x-11\right):3=3\)
\(\Leftrightarrow2x-11=9\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
Vậy x = 10.
d, \(\left(25-2x\right)\cdot3:5-32=42\)
\(\Leftrightarrow\)\(\frac{3\cdot\left(25-2x\right)}{5}=74\)
\(\Leftrightarrow3\cdot\left(25-2x\right)=370\)
\(\Leftrightarrow25-2x=\frac{370}{3}\)
\(\Leftrightarrow2x=-\frac{295}{3}\)
\(\Leftrightarrow x\approx49\)
Vậy \(x\approx49\) .
e, \(2\cdot3x=10\cdot312+8\cdot274\)
\(\Leftrightarrow6x=5312\)
\(\Leftrightarrow x=5312:6\approx885\)
Vậy \(x\approx885\) .
g, \(x-12=\left(-8\right)+\left(-17\right)\)
\(\Leftrightarrow x-12=-25\)
\(\Leftrightarrow x=-25+12=-13\)
Vậy x = -13.
h, \(7-2x=18-3x\)
\(\Leftrightarrow-2x+3x=18-7\)
\(\Leftrightarrow x=11\)
Vậy \(x=11\) .
i, \(3\cdot\left(x+5\right)-x-11=24\)
\(\Leftrightarrow3x+15-x-11=24\)
\(\Leftrightarrow2x=24+11-15\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\) .
19 tháng 4 2023
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
TT
1
11 tháng 8 2021
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
20 tháng 4 2020
a)
\(\left(4x-10\right)\cdot\left(24+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};-\frac{24}{5}\right\}\)
b)
\(\left(2x-5\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};\frac{2}{3}\right\}\)
c)
\(\left(2x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{1}{2};-\frac{1}{3}\right\}\)
d)
\(x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;\frac{1}{2}\right\}\)
e) \(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\)
Do \(x^2\ge0\) Nên \(x^2+4>0\)
\(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{-\frac{3}{5};1\right\}\)
....... Còn lại cứ cho mỗi thừa số = 0 rồi tìm x như bình thường thôi bạn
20 tháng 4 2020
1. (4x - 10)(24 + 5x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{-24}{5}\)}
2. (2x - 5)(3x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{2}{3}\)}
3. (2x - 1)(3x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{1}{2}\); \(\frac{-1}{3}\)}
4. x(x2 - 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy S = {0; 1; -1}
5. (5x + 3)(x2 + 4)(x - 1) = 0
VÌ x2 + 4 > 0 với mọi x nên
\(\Rightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{5}\\x=1\end{matrix}\right.\)
Vậy S = {\(\frac{-3}{5}\); 1}
6. (x - 1)(x + 2)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy S = {1; -2; -3}
7. (x - 1)(x + 5)(-3x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\frac{8}{3}\end{matrix}\right.\)
Vậy S = {1; -5; \(\frac{8}{3}\)}
Chúc bn học tốt!!
`1/8+(2x+5)/24=(3x+5)/2+x/6`
`=>3+2x+5=12(3x+5)+4x`
`=>2x+8=36x+60+4x`
`=>2x+8=40x+60`
`=>38x=-52`
`=>x=-26/19`
Vậy `x=-26/19`