35:0,5 +35 *99 -35
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a) \(S=1^5+3^5+....+75^5+99^5\)
\(\left(2a+1\right)^5-\left(2a+1\right)=2a\left(2a+1\right)\left(2a+2\right)\left[\left(2a+1\right)^2+1\right]\)
\(\left(2a+1\right)^5-\left(2a+1\right)=4a\left(2a+1\right)\left(a+1\right)\left[\left(2a+1\right)^2+1\right]⋮4\)
\(S=\left(1^5-1\right)+\left(3^5-3\right)+....+\left(75^5-75\right)+\left(99^5-99\right)+\left(1+3+5+...+75+99\right)\)
\(\Leftrightarrow\begin{matrix}1^5-1⋮4\\3^5-3⋮4\\5^5-5⋮4\\...........\\75^5-5⋮4\\99^5-99⋮4\end{matrix}\)
\(S_1=1+3+5+7+...+75+99=\frac{\left(1+75\right)\left[\frac{75-1}{2}+1\right]}{2}+99=38.38+96+3\)
\(\Rightarrow S_1:4\) dư 3
\(\Leftrightarrow S\) chia 4 dư 3
Sửa đề
\(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{-9}{35}\)
\(x+2=35\)
\(x=35-2\)
\(x=33\)
\(-17x=-17-\left(-34\right)\)
\(-17x=-17\)
\(x=-1\)
Bạn coi lại đề: $\frac{3}{69}$ không nằm trong các số hạng có quy luật.
`@` `\text {Ans}`
`\downarrow`
`a,`
\(x \div 0,1 + x \div 0,25 - 35 = 35\)
`\Rightarrow x * 10 + x * 4 - 35 = 35`
`\Rightarrow x*(10+4) - 35 = 35`
`\Rightarrow 14x - 35 = 35`
`\Rightarrow 14x = 70`
`\Rightarrow x=70/14 =5`
Vậy, `x=5`
`b,`
\(x \div 0,125 - x \div 0,5 = 108\)
`\Rightarrow x * 4/5 - x*2=108`
`\Rightarrow x*(4/5 - 2) = 108`
`\Rightarrow -6/5x = 108`
`\Rightarrow x=108 \div -6/5`
`\Rightarrow x=-90`
Vậy, `x=-90.`
a) x/0,1 + x/0,25 - 35 = 35
x . 10 + x . 4 = 35 + 35
14x = 70
x = 70 : 14
x = 5
=00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
A = 99 x2 + 33+333+3333+...+333...33 (100 chữ số 3)
đặt B = 33+333+3333+...+3333....33 = (99+999+9999+....+999...99): 3
= (102-1+103-1+104-1+....+ 10100-1):3
= (102+103+104+....+ 10100-99):3
đặt C= 102+103+104+....+ 10100
Cx 10 = 103+104+....+ 10100+10101
Suy ra C x9 = 10101- 102
Bạn tính tiếp nhé.
Ta có : \(\frac{x-99}{5}+\frac{x-99}{15}+\frac{x-99}{25}+\frac{x-99}{35}=0\)
\(\Rightarrow\left(x-99\right)\left(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\right)=0\)
Vì \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\ne0\)
Nên : x - 99 = 0
Suy ra : x = 99
Vậy x = 99
`4/15+35/99+11/15-2/7+64/99-5/7`
`=(4/15+11/15)+(35/99+64/99)-(2/7+5/7)`
`=15/15+99/99-7/7`
`=1+1-1`
`=2-1`
`=1`
= 70 + 3465 - 35
= 3500
35:0,5+35*99-35=35*2+35*99-35*1
= 35* (2+99-1)
= 35*100
=3500