Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,-4/7=x/21
-12/21 = x/21
x= -12
b,(x-3)/15=1/-5
x - 3 = -1/5 * 15
x - 3 = -3
x = 0
c,.(3x+8)/-12=-5/30
=> 3x + 8 = 2
=> 3x=-6
=>x=-2
a) 120 = 2^3*3*5
180 = 2^2*3^2*5
b)ƯCNN(120;180 ) = 2 *3*5=30
BCNN (120;180 ) = 2^3*3^2*5 = 360
2a) 3x - 12 = 27
3x = 27 - 12
3x = 15
x = 15:3
x = 5
b) theo đề bài ta có :
6 chia hết cho (x-1 )
=> x - 1 \(\in\) Ư(6)
mà Ư(6 ) = { 1;2;3;6;-1;-2;-3;-6 }
=> x - 1 = 1 => x = 2
x - 1 = 2 => x = 3
x - 1 = 3 => x = 4
x - 1 = 6 => x = 7
x - 1 = -1 => x = 0
x - 1 = -2 => x = -1
x - 1 = -3 => x = -2
x - 1 = -6 => x = -5
=> x \(\in\) { 2;3;4;7;0;-1;-2;-5 }
3.
Gọi số học sinh khối 6 trường đó có là a
theo đề bài ta có :
a chia hết cho 10;12;15
=> a \(\in\) BC (10;12;15)
Ta có :
10 = 2*5
12 = 2^2*3
15 = 3*5
=> BCNN (10;12;15 ) = 2^2 *3*5 = 60
=> BC (10;12;15 ) = B(60 ) = { 0;60;120;180;240;300;360 ;...}
Vì \(250\le a\le320\)
Nên a = 300
Vậy khối 6 trường đó có 300 học sinh
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
`Answer:`
a. \(x+12=3\Leftrightarrow x=3-12\Leftrightarrow x=-9\)
b. \(2x-15=21\Leftrightarrow2x=21+15\Leftrightarrow2x=36\Leftrightarrow x=36:2\Leftrightarrow x=18\)
c. \(13-3x=4\Leftrightarrow-3x=4-13\Leftrightarrow-3x=-9\Leftrightarrow x=-9:-3\Leftrightarrow x=3\)
d. \(2\left(x-2\right)+4=12\Leftrightarrow2x-4+4=12\Leftrightarrow2x=12\Leftrightarrow x=12:2\Leftrightarrow x=6\)
e. \(15-3\left(x-2\right)=21\Leftrightarrow15-3x+6=21\Leftrightarrow-3x=21-15-6\Leftrightarrow-3x=0\Leftrightarrow x=0\)
g. \(25+4\left(3-x\right)=1\Leftrightarrow25+12-4x=1\Leftrightarrow37-4x=1\Leftrightarrow-4x=-36\Leftrightarrow x=9\)
h. \(3x+12=2x-4\Leftrightarrow3x-2x=-4-12\Leftrightarrow x=-16\)
i. \(14-3x=\left(-x\right)+4\Leftrightarrow-3x+x=4-14\Leftrightarrow-2x=10\Leftrightarrow x=5\)
k. \(2\left(x-2\right)+7=x-25\Leftrightarrow2x-4+7=x-25\Leftrightarrow2x-x=-25-3\Leftrightarrow x=-28\)
a) \(-28-7|-3x+15|=-70\)
\(\Rightarrow7|-3x+15|=42\)
\(\Rightarrow|-3x+15|=6\)
\(\Rightarrow|3\left(5-x\right)|=6\)
\(\Rightarrow|3|.|5-x|=6\)
\(\Rightarrow3|5-x|=6\)
\(\Rightarrow|5-x|=2\)
\(\Rightarrow\orbr{\begin{cases}5-x=2\\5-x=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)
Vậy \(x\in\left\{3;7\right\}\)
b) \(|18-2|-x+5||=12\)
\(\Rightarrow\orbr{\begin{cases}18-2|-x+5|=12\\18-2|-x+5|=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2|5-x|=6\\2|5-x|=30\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}|5-x|=3\left(1\right)\\|5-x|=15\left(2\right)\end{cases}}\)
Từ \(\left(1\right):\Rightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)
Từ \(\left(2\right):\Rightarrow\orbr{\begin{cases}5-x=15\\5-x=-15\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-10\\x=20\end{cases}}\)
Vậy \(x\in\left\{2;8;-10;20\right\}\)
c) \(12-2\left(-x+3\right)^2=-38\)
\(\Rightarrow2\left(3-x\right)^2=50\)
\(\Rightarrow\left(3-x\right)^2=100\)
\(\Rightarrow\orbr{\begin{cases}3-x=10\\3-x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-7\\x=13\end{cases}}\)
Vậy \(x\in\left\{-7;13\right\}\)
d) \(-20+3\left(2x+1\right)^3=-101\)
\(\Rightarrow3\left(2x+1\right)^3=-81\)
\(\Rightarrow\left(2x+1\right)^3=-27\)
\(\Rightarrow2x+1=-3\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
Trả lời:
a, -28 - 7| -3x + 15 | = -70
=> 7| -3x + 15 | = 42
=> | -3x + 15 | = 6
=> -3x + 15 = 6 hoặc -3x + 15 = -6
=> -3x = -9 -3x = -21
=> x = 3 x = 7
Vậy x = 3; x = 7
b, | 18 - 2 | -x + 5 || = 12
=> 18 - 2| -x + 5 | = 12 hoặc 18 - 2| -x + 5 | = -12
=> 2 | -x + 5 | = 6 hoặc 2 | -x + 5 | = 30
=> | -x + 5 | = 3 hoặc | -x + 5 | = 15
=> -x + 5 = 3 hoặc -x + 5 = -3 hoặc -x + 5 = 15 hoặc -x + 5 = -15
=> x = 2 x = 8 x = -10 x = 20
Vậy x \(\in\){ 2; 8; -10; 20 }
c, 12 - 2.( -x + 3 )2 = -38
=> 2.( -x + 3 )2 = 50
=> ( -x + 3 )2 = 25
=> -x + 3 = 5 hoặc -x + 3 = -5
=> x = -2 x = 8
Vậy x = -2; x = 8
d, -20 + 3.( 2x + 1 )3 = -101
=> 3.( 2x + 1)3 = -81
=> ( 2x + 1 )3 = -27
=> 2x + 1 = -3
=> 2x = -4
=> x = -2
Vậy x = -2
=> x = 1