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a) 1 + 2 + 3 + .... + x = 120
<=> x(x + 1) : 2 = 120
<=> x(x + 1) = 240
<=> x(x + 1) = 15.16
<=> x = 15 (Vì x \(\inℕ\))
b) (x + 1) + (x + 2) + (x + 3) + .... + (x + 100) = 12650
=> (x + x + x + .... + x) + (1 + 2 + 3 + .... +100) = 12650 (100 hạng tử x)
=> 100x + 100.(100 + 1) : 2 = 12650
=> 100x + 5050 = 12650
<=> 100x = 7600
<=> x = 76
Vậy x = 76 là giá trị cần tìm
Đề bài hơi lằng nhằng nên có 2 kiểu nhé bạn :
Nếu đề bài là 2x + 2x +1 +2x+2+2x + 3 =120
2x ( 1+2+3)=120
2x 6 = 120
2x= 120 : 6
2 mũ x =20
x thuộc rỗng :v
Đề kiểu 2 :
Nếu đề bài là 2x + 2x+1 +2x+2 2x+3 =120
Thì là : 2x.1 + 2x . 21+2x . 22 2x . 23 =120
2x ( 1+2+4+8) =120
2 mũ x . 15 =120
2 mũ x = 8
x= 3
mình bạn nhé :)
a) \(2^{x-1}+2^{x+1}+2^{x+2}=104\)
=> \(2^{x-1}+2^x\cdot2+2^x\cdot2^2=104\)
=> \(2^x:2+2^x\cdot\left(2+2^2\right)=104\)
=> \(2^x\cdot\frac{1}{2}+2^x\cdot6=104\)
=> \(2^x\cdot\left(\frac{1}{2}+6\right)=104\Rightarrow2^x=104:\left(\frac{1}{2}+6\right)=104:\frac{13}{2}=16\)
=> \(x=4\)
a) 13 x ( x +1 ) =143
x+1= 143 : 13
x+1= 11
x= 11-1
x= 10
b) ( x - 2 ) : 7 = 12
x-2= 12 x 7
x-2=84
x=84+2
x=86
c) 120 : ( x - 3 )= 8
x-3=120:8
x-3=15
x=15+3
x=18
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=3^x\left(3^1+3^2+3^3+3^4\right)+...+3^{x+96}\left(3^1+3^2+3^3+3^4\right)=3^x.120+3^{x+4}.120+...+3^{x+96}.120=120\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
a) x - 120: 30 = 40
x -40 =40
x =40+40
x =80
b) (x + 120) : 20 = 8
(x+ 120) = 8x20
x+120 =160
x = 160-120
x = 40
c) (x + 5). 3 = 300
x+5=300:3
x+5=100
x=100-5
x=95
d) x.2 + 21 : 3= 27
x.2 +7=27
x.2 = 27-7
x.2= 20
x=20:2
x=10
\(a) \frac{4}{5}.x=\frac{8}{35}\)
\(\implies x= \frac{8}{35}:\frac{4}{5}\)
\(\implies x=\frac{8}{35}.\frac{5}{4}\)
\(\implies x=\frac{2}{7}\). Vậy \(x=\frac{2}{7}\)
\(b) \frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(\implies \frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(\implies \frac{3}{5}x=\frac{9}{14}\)
\(\implies x=\frac{9}{14}:\frac{3}{5}\)
\(\implies x=\frac{9}{14}.\frac{5}{3} \)
\(\implies x=\frac{15}{14}\). Vậy \(x=\frac{15}{14}\)
\(c) x-25\% x=0,5\)
\(\implies 75\% x=0,5\)
\(\implies \frac{3}{4}x=\frac{1}{2}\)
\(\implies x=\frac{1}{2}:\frac{3}{4}\)
\(\implies x=\frac{1}{2}.\frac{4}{3}\)
\(\implies x=\frac{2}{3}\). Vậy \(x=\frac{2}{3}\)
\(d) (\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
Có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\implies 1+x:\frac{1}{3}=-4\)
\(\implies x:\frac{1}{3}=-5\)
\(\implies x=-5.\frac{1}{3}\)
\(\implies x=\frac{-5}{3}\). Vậy \(x=\frac{-5}{3}\)
~ Hok tốt a~
\(\frac{4}{5}.x=\frac{8}{35}\) \(\frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(x=\frac{8}{35}:\frac{4}{5}\) \(\frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(x=\frac{8}{35}.\frac{5}{4}\) \(\frac{3}{5}x=\frac{9}{14}\)
\(x=\frac{2}{7}\) \(x=\frac{9}{14}:\frac{3}{5}\)
Vậy \(x=\frac{2}{7}\) \(x=\frac{9}{14}.\frac{5}{3}\)
\(x-25\%x=0.5\) \(x=\frac{15}{14}\)
\(x-\frac{1}{4}x=\frac{1}{2}\) Vậy \(x=\frac{15}{14}\)
\(x\left(1-\frac{1}{4}\right)=\frac{1}{2}\) \(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(x=\frac{1}{2}:\frac{3}{4}\) \(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\) \(x=\frac{2}{3}\) \(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
Vậy \(x=\frac{2}{3}\) \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=\left(-4\right)-1\)
\(x:\frac{1}{3}=-5\)
\(x=\left(-5\right).\frac{1}{3}\)
\(x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}\)
(x + 1) + (x + 2) + (x + 3) + ... + (x + 102) = 5559
x + 1 + x + 2 + x + 3 + ... + x + 102 = 5559
102x + (1 + 2 + 3 + ... + 102) = 5559
102x + (102 + 1) . 102 : 2 = 5559
102x + 5253 = 5559
102x = 5559 - 5253
102x = 306
X = 306 : 102
X = 3