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Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
Đặt A = 1.2 + 2.3 + 3.4 + ...... + 99.100
3A=1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .....+99.100.101
3A=99.100.101
A=99.100.101/3=333300
41 - (2x - 5) = 18
=> 2x - 5 = 41 - 18
=> 2x - 5 = 23
=> 2x = 23 + 5
=> 2x = 28
=> x = 28 : 2
=> x = 14
Vậy x = 14
41-(2x-5)=18
2x-5=41-18
2x-5=23
2x=23+5
2x=28
x=28:2=14
vậy x=14
a) \(-39-\left(15+x\right)=18\)
\(-39-15-x=18\)
\(-54-x=18\)
\(x=-54-18\)
\(x=-72\)
vậy \(x=-72\)
b) \(31-\left(13+x\right)=18.\left(-4\right)\)
\(31-13-x=-72\)
\(18-x=-72\)
\(x=18+72\)
\(x=90\)
vậy \(x=90\)
c) \(58-\left(47+3x\right)=-18-2x\)
\(58-47-3x+2x=-18\)
\(11-x=-18\)
\(x=11+18\)
\(x=29\)
vậy \(x=29\)
d) \(37-\left(115+x\right)=188\)
\(37-115-x=188\)
\(-78-x=188\)
\(x=-78-188\)
\(x=-266\)
vậy \(x=-266\)
a)
\(-39-\left(15+x\right)=18\)
\(\Rightarrow15+x=-39-18\)
\(\Rightarrow15+x=-57\)
\(\Rightarrow x=-57-15\)
\(\Rightarrow x=-72\)
Vậy \(x=-72\)
b)
\(31-\left(13+x\right)=18.\left(-4\right)\)
\(\Rightarrow31-\left(13+x\right)=-72\)
\(\Rightarrow13+x=31-\left(-72\right)\)
\(\Rightarrow13+x=103\)
\(\Rightarrow x=103-13\)
\(\Rightarrow x=90\)
Vậy \(x=90\)
=>(2x-7)2=9
=>2x-7 =3 hoặc 2x-7=-3
=>2x=10 hoặc 2x=4
=>x=5 hoặc x=2
Vậy ....................
\(2\cdot\left(2x-7\right)^2=18\)
\(\Rightarrow\left(2x-7\right)^2=9\)
\(\Rightarrow\hept{\begin{cases}2x-7=3\\2x-7=-3\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=10\\2x=4\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=2\end{cases}}\)
(-2x-7)-(-5x+18)=35
=>-2x-7+5x-18=35
=>3x-25=35
=>3x=35+25
=>3x=60
=>x=60:3
=>x=20
(-2x-7)-(-5x+18)=35
-2x-7+5x-18=35
(-2x+5x)-(7+18)=35
3x-25=35
3x=35+25
3x=60
x=60:3
x=20
Vậy x=20
3x+2 - 2.3x = 63
=> 3x . 32 - 2.3x = 63
=> 3x(32 - 2) = 63
=> 3x(9 - 2) = 63
=> 3x . 7 = 63
=> 3x = 9
=> 3x = 32 => x = 2
\(3^{x+2}-2.3^x=63\)
\(3^x.3^2-2.3^x=63\)
\(3^x.\left(9-2\right)=63\)
\(3^x.7=63\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
A = \(\left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+...+\left(1+\frac{1}{99.100}\right)\)(99 số hạng)
= \(\left(1+1+....+1\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)(99 số hạng 1)
= \(99.1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(99+\left(1-\frac{1}{100}\right)=99+\frac{99}{100}=99,99\)
2.32x = 18
32x = 18 : 2 = 9
Có : 32 = 9
⇒2x = 2
⇒x = 1
Vậy x = 1
\(2.3^x=18\)
\(3^x=18:2\)
\(3^x=9\)
`3^x=3^2`
`x=2`