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\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}-\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+...+\dfrac{7}{23.30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{23}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\right)\)
\(=\dfrac{1}{7}\left(\dfrac{77}{240}-\dfrac{7}{15}\right)=\dfrac{1}{7}.\left(-\dfrac{7}{48}\right)=-\dfrac{1}{48}\)
Nếu phân số thứ 2 là \(\frac{1}{10.17}\) thì làm như vậy nè
\(\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{73.80}-\frac{1}{2.9}-\frac{1}{9.16}-\frac{1}{16.23}-\frac{1}{23.30}\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}\right)-\left(\frac{1}{2.9}+\frac{1}{9.16}+\frac{1}{16.23}+\frac{1}{23.30}\right)\)
= \(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}\right)-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
= \(\frac{1}{7}.\frac{77}{240}-\frac{1}{7}\left(\frac{1}{2}-\frac{1}{30}\right)=\frac{1}{7}.\frac{77}{240}-\frac{1}{7}.\frac{7}{15}\)
= \(\frac{11}{240}-\frac{1}{15}\)
= \(-\frac{1}{48}\)
Ta có:
P=\(\frac{1}{3.10}\)+\(\frac{1}{10.17}\)+\(\frac{1}{17.24}\)+......+\(\frac{1}{73.80}\)-\(\frac{1}{2.9}\)-\(\frac{1}{9.16}\)-\(\frac{1}{16.23}\)-\(\frac{1}{23.30}\))
P=\(\frac{1}{7}\)\(\times\)(\(\frac{7}{3.10}\)+\(\frac{7}{10.17}\)+\(\frac{7}{17.24}\)+......\(\frac{7}{73.80}\)-\(\frac{7}{2.9}\)-\(\frac{7}{9.16}\)-\(\frac{7}{16.23}\)-\(\frac{7}{23.30}\))
P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{17}\)+.....+\(\frac{1}{73}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)-\(\frac{1}{9}\)-......-\(\frac{1}{23}\)-\(\frac{1}{30}\))
P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\))-\(\frac{1}{7}\)(\(\frac{1}{2}\)-\(\frac{1}{30}\))
P=\(\frac{1}{7}\)\(\times\)(\(\frac{1}{3}\)-\(\frac{1}{80}\)-\(\frac{1}{2}\)+\(\frac{1}{30}\))
P=\(\frac{-7}{336}\)
Bài này mk ko tính máy tính nên ko chắc đâu
taị mk ko tính máy tính lên sai.
bn thông cảm nha. thường ngày hay dùng máy tính quá nên tính sai thì bn thông cảm
\(\dfrac{x}{-5}=\dfrac{y}{4}\)
⇒\(\dfrac{3x}{-15}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{-15}=\dfrac{5y}{20}=\dfrac{3x-5y}{-15-35}=\dfrac{-210}{-35}=6\)
⇒\(\left\{{}\begin{matrix}x=6.-5=-30\\y=6.4=24\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{-5}=\dfrac{y}{4}=\dfrac{3x-5y}{3\cdot\left(-5\right)-5\cdot4}=\dfrac{-210}{-35}=6\)
Do đó: x=-30; y=24
2210 và 5140
Ta có :
\(2^{210}=\left(2^3\right)^{70}=8^{70}\)
\(5^{140}=\left(5^2\right)^{70}=25^{70}\)
Vì \(8^{70}< 25^{70}\)nên \(2^{210}< 5^{140}\)
Ta có : 2210 = (23)70 = 870
5140 = (52)70 = 2570
Mà : 870 < 2570
Nên : 2210 < 5140
\(2^{210}=\left(2^3\right)^{70}=8^{70}\)
\(5^{140}=\left(5^2\right)^{70}=25^{70}\)
870<2570 <=> 2210<5140
\(C=\dfrac{5122512}{2^2}-512\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}\right)\)
Đặt BT trong ngoặc đơn là B
\(\Rightarrow2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
\(B=2B-B=\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\)
\(\Rightarrow C=\dfrac{5120512+2000}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)
\(=\dfrac{512.10001+2^2.500}{2^2}-512\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)
\(=\dfrac{2^9.10001+2^2.500}{2^2}-2^9\left(\dfrac{1}{2^2}-\dfrac{1}{2^{10}}\right)=\)
\(=2^7.10001+500-2^7+\dfrac{1}{2}=\)
\(=2^7.10000+500+0,5=1280000+500+0,5=1280500,5\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`210 \div x - 1/2 = 20,5`
`=> 210 \div x = 20,5 + 1/2`
`=> 210 \div x =21`
`=> x = 210 \div 21`
`=> x = 10`
Vậy, `x = 10.`
`b)`
`7 * 3^x + 20*3^x = 3^25`
`=> 3^x * (7+20) = 3^25`
`=> 3^x * 27 = 3^25`
`=> 3^x * 3^3 = 3^25`
`=> 3^x = 3^25 \div 3^3`
`=> 3^x = 3^22`
`=> x = 22`
Vậy, `x = 22.`
a) \(210:x-\dfrac{1}{2}=20,5\)
\(\Rightarrow210:x=20,5+\dfrac{1}{2}\)
\(\Rightarrow210:x=21\)
\(\Rightarrow x=\dfrac{210}{21}\)
\(\Rightarrow x=10\)
b) \(7\cdot3^x+20\cdot3^x=3^{25}\)
\(\Rightarrow3^x\cdot\left(7+20\right)=3^{25}\)
\(\Rightarrow3^x\cdot27=3^{25}\)
\(\Rightarrow3^x\cdot3^3=3^{25}\)
\(\Rightarrow3^{x+3}=3^{25}\)
\(\Rightarrow x+3=25\)
\(\Rightarrow x=25-3\)
\(\Rightarrow x=22\)
Ta có : 2 + 4 + 6 +......+ 2x = 210
=> \(\frac{\left[\left(2x-2\right):2+1\right]\left(2x+2\right)}{2}=210\)
=> \(\frac{\left[x-1+1\right]\left(2x+2\right)}{2}=210\)
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
=> x (x + 1) = 14.15
=> x = 14
Vậy x = 14 .
=10000
học tốt
=1024.9+1024-8.30
=1024.(9-8).30
=1024.30
=30720