a)-2.x=-16
b)x.-12=-42
c)-17-x=13
d)x+5=-14
e)x.(-2)=10
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c: Ta có: 11+x:5=13
\(\Leftrightarrow x:5=2\)
hay x=10
d: Ta có: \(13+2\left(x+1\right)=15\)
\(\Leftrightarrow2x+2=2\)
\(\Leftrightarrow2x=0\)
hay x=0
e: Ta có: 2x+21=41
\(\Leftrightarrow2x=20\)
hay x=10
f: Ta có: \(12+3\left(x-2\right)=60\)
\(\Leftrightarrow3\left(x-2\right)=48\)
\(\Leftrightarrow x-2=16\)
hay x=18
g: Ta có: \(24x-11\cdot13=11\cdot11\)
\(\Leftrightarrow24x=11\cdot24\)
hay x=11
h: Ta có: \(17-\left(x-4\right):2=3\)
\(\Leftrightarrow\left(x-4\right):2=14\)
\(\Leftrightarrow x-4=28\)
hay x=32
c: 90\(⋮\)x-2
=>\(x-2\in\){1;-1;2;-2;3;-3;5;-5;6;-6;9;-9;10;-10;15;-15;18;-18;30;-30;45;-45;90;-90}
=>x\(\in\){3;1;4;0;5;-1;7;-3;8;-4;11;-7;12;-8;17;-13;20;-16;32;-28;47;-43;92;-88}
mà 10<=x<=90
nên \(x\in\left\{11;12;17;20;32;47\right\}\)
d: \(x⋮12\)
=>\(x\in\left\{0;12;24;36;48;60;72;...\right\}\)
mà 30<=x<=60
nên \(x\in\left\{36;48;60\right\}\)
a: x-140:35=270
=>x-4=270
=>x=270+4=274
b: \(\left(3x-1\right)^2-3^2=4^2\)
=>(3x-1)2-9=16
=>\(\left(3x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{2}{5}\)
\(x=\dfrac{5}{10}-\dfrac{4}{10}\)
\(\Rightarrow x=....\)
\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{5-4}{10}=\dfrac{1}{10}\)
\(b,x-\dfrac{2}{5}=\dfrac{1}{7}\\ \Rightarrow x=\dfrac{1}{7}+\dfrac{2}{5}=\dfrac{5+14}{35}=\dfrac{19}{35}\)
\(c,x\cdot\dfrac{3}{4}=\dfrac{9}{20}\\ \Rightarrow x=\dfrac{9}{20}:\dfrac{3}{4}=\dfrac{9}{20}\cdot\dfrac{4}{3}=\dfrac{3\cdot1}{5\cdot1}=\dfrac{3}{5}\)
\(d,x:\dfrac{1}{7}=14\\ \Rightarrow x=14\cdot\dfrac{1}{7}=\dfrac{14}{7}=2\)
\(e,\dfrac{2}{3}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{2}{3}-\dfrac{1}{5}=\dfrac{10-3}{15}=\dfrac{7}{15}\)
\(f,\dfrac{4}{15}:x=\dfrac{12}{25}\\ \Rightarrow x=\dfrac{4}{15}:\dfrac{12}{25}=\dfrac{4}{15}\cdot\dfrac{25}{12}=\dfrac{1\cdot5}{3\cdot3}=\dfrac{5}{9}\)
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
( Mik làm mấy phần mà bạn dưới chưa làm)
11) xy+x+y=9
\(\Leftrightarrow\) xy+x+y+1=9+1
\(\Leftrightarrow\left(xy+x\right)+\left(y+1\right)\)=10
\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=10\)
\(\Leftrightarrow\) (x+1)(y+1)=10=1.10=10.1=-1.-10=-10.-1=2.5=5.2=-2.-5=-5.-2
\(\Rightarrow\) TH1: x+1=1 ; y+1=10
\(\Leftrightarrow x=0;y=9\)
TH2: x+1=10;y+1=1
\(\Leftrightarrow\)x=9;y=0
TH3: x+1=-1;y+1=-10
\(\Leftrightarrow\) x=-2;y=-11
...........
Vậy:........
( Bạn tự làm nốt chứ dài quá, mik chỉ hướng dẫn cách làm bài thôi)
1) -x = -7
=> x = 7
2) - x = 17
=> x = - 17
3) |x| = 17
=> x = ±17
4) -(-x) = |-17|
=> x = 17
5) - 19 - x = 17
=> - x = 17 + 19
=> x = - 36
6) - 19 - x = - 17
=> - x = - 17 + 19
=> -x = 2
=> x = - 2
7) - 5 - (10 - x) = 7
=> - 5 - 10 + x = 7
=> - 15 + x = 7
=> x = 7 + 15
=> x = 22
8) |x + 3| + 7 = 12
=> |x + 3| = 12 - 7
=> |x + 3| = 5
=> x + 3 = 5 hoặc x + 3 =- 5
=> x = 2 hoặc x = - 8
9) 2 - |x - 2| = x
=> - |x - 2| - x = - 2
TH1: x >= 2
- (x - 2) - x = - 2
=> - x + 2 - x =- 2
=> - 2x = - 4
=> x = 2 (nhận)
TH2: x < 2
-[-(x - 2)] - x = - 2
=> x - 2 - x = - 2
=> 0x = 0 (vô số nghiệm)
a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
a)-2.x=-16
x= -16:(-2)=8
Vậy....
b)x.-12=-42
x=-42:(-12)=3,5
Vậy....
c)-17-x=13
x= -17-13=-30
Vậy....
d)x+5=-14
x=-14-5=-19
Vậy....
e)x.(-2)=10
x=10:(-2)=-5Vậy....a, -2x=-16
x= (-16):(-2)
x= 8
vậy...
b,
x.(-12)=-42
x= (-42):(-12)
x= 7/2
Vậy...
c,
-17-x=13
x= (-17)-13
x=(-30)
Vậy....
d, x+5=-14
x= -14-5
x=-19
Vậy....
e, x.(-2)=10
x= 10 : (-2)
x= -5
Vậy...