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-7-2x = 37-(-26)
Suy ra -7-2x=11
Suy ra -2x=11-7
Suy ra -2x=4
Suy ra 2x= -4
Suy ra x = -4 :2 =-2
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 \(x=-\dfrac{154}{29}\)
Trả lời :.........................
\(20⋮x\Rightarrow x\inƯ\left(20\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
Hk tốt.............................
câu 1: x thuộc {1;2;4;5;10;20}
câu 2: 2x+9=2019
=>2x=2010
=>x=1005
câu 3: 3(x-5)-11=37
=>3(x-5)=48
=>x-5=16
=>x=21
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
Ta có -2-(x-17)=34-(-x+25)
-2-x+17=34+x+25
-2 +(-x)+17+(-x)=34+25
[(-x)+(-x)]+[(-2)+17]=29
(-x).2+15=29
(-x).2=29-15
(-x).2=14
(-x)=14:2
(-x)=7
x=-7
\(11\times\left[\left(2x-1^3\right)-1\right]+37=1401\)
\(11\times\left[\left(2x-1\right)-1\right]=1401-37\)
\(\left(2x-1\right)-1=1364\div11\)
\(2x-1=124+1\)
\(2x=125+1\)
\(2x=126\)
\(\Rightarrow x=63\)
Chắc chắn đúng 100% luôn !
11[(2x-13)-1]+37=1401
=> 11[(2x-1)-1]=1401-37
=> 11[(2x-1)-1]=1364
=> (2x-1)-1=1364:11
=> (2x-1)-1=124
=> 2x-1=124+1
=> 2x-1=125
=> 2x=125+1
=> 2x=126
=> x=126:2
=> x=63