Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
45 - 3 x (x+2) = 8 x 3 = 24
3 x (x+2) = 45 - 24 = 21
x+2 = 21 : 3 = 7
x = 7-2 = 5
k mk nha
3 x (x+2) = 45 - 24 = 21
x+2 = 21 : 3 = 7
x = 7-2 = 5
chúc bn hok tốt @_@
3*2^x-3=45
3*2^x=45+3=48
2^x=48:3=16
2^x=2^4
=> x=4
Vậy x=4
3.2x-3=45
\(\Rightarrow\)3.2x=45+3
\(\Rightarrow\)3.2x=48
\(\Rightarrow\)2x=48:3
\(\Rightarrow\)2x=16
\(\Rightarrow\)2x=24
\(\Rightarrow\)x=4
\(3.2^x-3=45\)
\(3.2^x=45+3\)
\(3.2^x=48\)
\(2^x=48:3\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
3 . 2x - 3 = 45
<=> 3 . 2x = 48
<=> 2x = 16
<=> 2x= 24
<=> x = 4(t/m)
-45 : 5 (-3-2x)=3
<=> -9(-3-2x)=3
<=>-3-2x=-1/3
<=> 2x = -8/3
<=> x = -4/3
Vậy ...
\(a,14-5+x=30\)
\(\Rightarrow9+x=30\Rightarrow x=30-9\)
\(\Rightarrow x=21\)
\(b,45-\left(3+x\right)=14\)
\(\Rightarrow3+x=45-14\)
\(\Rightarrow3+x=31\Rightarrow x=31-3\)
\(\Rightarrow x=28\)
\(c,18+\left(-3+x\right)-\left(44-x\right)=55\)
\(\Rightarrow18-3+x-44+x=55\)
\(\Rightarrow2x-29=55\Rightarrow2x=55+29\)
\(\Rightarrow2x=84\Rightarrow x=84:2\)
\(\Rightarrow x=42\)
\(d,\left(x-45\right).27=0\)
\(\Rightarrow x-45=0\Rightarrow x=45\)
7 + 2x = 45 : 3
7 + 2x = 15
2x = 15 - 7
2x = 8
x = 8 : 2
x = 4
Vậy x = 4
a) Dãy trên có số số hạng là :
( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x
=> 1 + 2 + 3 + 4 + ... + x = 45
=> ( x + 1 ) . x : 2 = 45
=> ( x + 1 ) . x = 90
=> ( x + 1 ) . x = 10 . 9
=> x = 9
Vậy x = 9
a) Dãy trên có số số hạng là :
( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x
=> 1 + 2 + 3 + 4 + ... + x = 45
=> ( x + 1 ) . x : 2 = 45
=> ( x + 1 ) . x = 90
=> ( x + 1 ) . x = 10 . 9
=> x = 9
Vậy x = 9
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
2/3+x=-45%
2/3+x=-9/20
x=-9/20-2/3
x=-67/60
Vậy x= -67/60
2/3 + x = -45%
2/3 + x = -9/20
x = -9/20 - 2/3
x = -67/60
=^_^=
k mình nha