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1) \(3\left(5x-2\right)-9=6\)
\(15x-6=6+9\)
\(15x=15+6\)
\(15x=21\)
\(x=\frac{7}{5}\)
2) \(4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\left(3x-2\right)\left(4+7\right)=22\)
\(\left(3x-2\right)\cdot11=22\)
\(\left(3x-2\right)=2\)
\(3x=4\)
\(x=\frac{4}{3}\)
3) \(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy,............
\(1)3\left(5x-1\right)-9=6\)
\(\Rightarrow3\left(5x-2\right)=15\)
\(\Rightarrow5x-2=5\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\)
\(2)4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\Rightarrow11\left(3x-2\right)=22\)
\(\Rightarrow33x-22=22\Rightarrow33x=44\Rightarrow x=\frac{4}{3}\)
\(3)2x^2-6x=0\)
\(\Rightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
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
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 \(x=10\) .
2:
a: =>2(x+1)=26
=>x+1=13
=>x=12
b: =>(6x)^3=125
=>6x=5
=>x=5/6(loại)
c: =>\(7\cdot3^x\cdot\dfrac{1}{3}+11\cdot3^x\cdot3=318\)
=>3^x=9
=>x=2
d: -2x+13 chia hết cho x+1
=>-2x-2+15 chia hết cho x+1
=>15 chia hết cho x+1
=>x+1 thuộc {1;3;5;15}
=>x thuộc {0;2;4;14}
e: 4x+11 chia hết cho 3x+2
=>12x+33 chia hết cho 3x+2
=>12x+8+25 chia hết cho 3x+2
=>25 chia hết cho 3x+2
=>3x+2 thuộc {1;-1;5;-5;25;-25}
mà x là số tự nhiên
nên x=1
1:
a: Đặt A=2^2024-2^2023-...-2^2-2-1
Đặt B=2^2023+2^2022+...+2^2+2+1
=>2B=2^2024+2^2023+...+2^3+2^2+2
=>B=2^2024-1
=>A=2^2024-2^2024+1=1
c: \(=\dfrac{3^{12}\cdot2^{11}+2^{10}\cdot3^{12}\cdot5}{2^2\cdot3\cdot3^{11}\cdot2^{11}}=\dfrac{2^{10}\cdot3^{12}\left(2+5\right)}{2^{13}\cdot3^{12}}\)
\(=\dfrac{7}{2^3}=\dfrac{7}{8}\)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
a) x - 14 = 3x + 18
=> x - 3x = 18 + 14
=> -2x = 32
=> x = 32 : (-2)
=> x = -16
b) 2(x - 5) - 3(x - 4) = -6 + 15.(-3)
=> 2x - 10 - 3x + 12 = -6 - 45
=> -x + 2 = -51
=> -x = -51 - 2
=> -x = -53
=> x = 53
c) (x + 7)(x - 9) = 0
=> \(\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-7\\x=9\end{cases}}\)
Vậy ...
d) |2x - 5| - 7 = 22
=> |2x - 5| = 22 + 7
=> |2x - 5| = 29
=> \(\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\)
=> \(\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\)
=> \(\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)
Vậy ...
- a\(x-14=3x+18\Rightarrow3x-x=-14-18=-32\Rightarrow x=-16\)
- b, \(2\left(x-5\right)-3\left(x-4\right)=-6+15.\left(-3\right)\Rightarrow2x-10-3x+12=-6-45\)
- \(\Rightarrow-x+2=-51\Rightarrow-x=-53\Rightarrow x=53\)
- c.\(\left(x+7\right).\left(x-9\right)=0\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
- d.Bạn tự làm nhé
d) \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)
Vậy x = 17 hoặc x = -12
a/ \(51-(-12+3x)=27\)
\(\Leftrightarrow51+12-27-3x=0\Leftrightarrow36=3x\Leftrightarrow x=\frac{36}{3}=12\)
KL:........
b/ $-x + 21=15+ 2x$
\(\Leftrightarrow2x+x=21-15\Leftrightarrow2x=6\Leftrightarrow x=3\)
KL: ...........
c) $7.(x-9)-5(6-x)=-6+11.x$
\(\Leftrightarrow7x-63-30+5x=-6+11x\Leftrightarrow7x+5x-11x=-6+63+30\Leftrightarrow x=87\)
KL:............
d) $(x-3).(x^2 + 2)=0$
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x\in\varnothing\end{matrix}\right.\)\(\Leftrightarrow x=3\)
e) $|2x-7|-22=-13$
\(\Leftrightarrow\left|2x-7\right|=9\Leftrightarrow\left[{}\begin{matrix}2x-7=9\\2x-7=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
KL: ...........
f) $(2x - 1)^3=-125$
\(\Leftrightarrow\left(2x-1\right)^3=\left(-5\right)^3\Leftrightarrow2x-1=-5\Leftrightarrow x=-2\)
KL: ...........
d) |2x – 5| –7 = 22
| 2x -5 | = 22+7
| 2x -5 | = 29
TH1: 2x-5 = -29
2x = -29+5
2x= -24
x= -24:2
x= -12
TH2: 2x -5 =29
2x = 29+5
2x= 34
x= 34:2
x= 17
Vậy...
7(3x+2)-3(2x-3)=-22
21x+14-6x+9=-22
21x-6x=-22-14-9
15x=-45
x=-45:15
x=-3
Vậy x=-3
\(7.\left(3.x+2\right)-3.\left(2.x-3\right)=-22\)
\(7.3.x+7.2-3.2.x+3.3=-22\)
\(21.x+14-6.x+9=-22\)
\(\left(21.x-6.x\right)+\left(14+9\right)=-22\)
\(15.x+23=-22\)
\(15.x=-22-23\)
\(15.x=-45\)
\(x=-45:15\)
\(x=-3\)