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a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
\(\)1)A={xEN*/x\(\ge\)6}
2) a)\(\Rightarrow\left(x+15\right).9=45\)
\(\Rightarrow x+15=45:9=5\)
\(\Rightarrow x=5-15\)
\(\Rightarrow x=-10\)
b) \(\Rightarrow10+2x=20\)
\(\Rightarrow2x=20-10=10\)
\(\Rightarrow x=10:2=5\)
tíc mình nha
a) x3 _ 11 = 4100 : 4 98
x3 - 11 = 42
x3 - 11 = 16
x3 = 16 + 11
x3 = 27
x3 = 33
x = 3
b) 102 + 2 . ( x - 4 ) = 120
100 + 2. ( x - 4 ) = 120
2. ( x - 4 ) = 120 - 100
2. ( x - 4 ) = 20
x - 4 = 20 : 2
x - 4 = 10
x = 10 + 4
x = 14
c) 3x+1+5.3x = 102 - 22.7
3x+1+5.3x = 102 - 4.7
3x+1+5.3x = 102 - 28
3x+1+5.3x = 100 - 28
3x+1+5.3x = 72
3x ( 3 + 5 ) = 72
3x . 8 = 72
3x = 72 : 8
3x = 9
3x = 32
x = 2
a) x3 -11=4100:498 b)102+2.(x-4)=120
x3 -11=42 2.(x-4)=120-102
x3 -11=16 2.(x-4)=20
x3 =16+11 (x-4)=20:2
x3 =27 x-4=10
Vì 33=27 nên x=3 x =10+4
x =14
a) (7x - 11)3 = 25.52 + 200
=> (7x - 11)3 = 1000
=> (7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> x = 21 : 7
=> x = 3
b) x10 = 1x
=> x10 = 1
=> x10 = 110
=> x = 1
c) x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy...
a) \(7x+x=3^2.10+6\) \(\Rightarrow8x=96\) \(\Rightarrow x=\frac{96}{8}=12\)
b) \(120-2\left(x-1\right)=10^{10}:10^8\) \(\Rightarrow120-2\left(x-1\right)=10^{10-8}\)
\(120-2\left(x-1\right)=100\)
\(2\left(x-1\right)=120-100=20\)
\(x-1=\frac{20}{2}=10\)
\(\Rightarrow x=10+1=11\)
a) <=> 8x= 96 => x=12
b) 120- 2(x-1) = 102
<=> 2(x-1)=120 - 102= 20
<=>x-1= 10 Suy ra: x=10+1=11