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a,
Có: n/n+1 = n+1-1/n+1 = 1-(1/n+1)
n+2/n+3 = n+3-1/n+3 = 1-(1/n+3)
Vì 1/n+1 > 1/n+3
=> 1-(1/n+1) < 1-(1/n+3) hay n/n+1 < n+2/n+3
b,
giả sử n/n+3 < n-1/n+4
<=> n(n+4) < (n+3)(n-1)
<=> n^2 + 4n < n^2 + 2n - 3
<=> 2n < -3 (sai)
vậy n/n+3 > n-1/n+4
c) \(\frac{n}{2n+1}\)= \(\frac{3n}{6n+3}\)< \(\frac{3n+1}{6n+3}\)
a,
53.(-15) + (-15).47
= (-15).(53 + 47)
= -15.100
= -1500
b,
-43.92 - 46.27 + 46.41
= -43.92 - 46.(27 - 41)
= -86.46 - 46.(-14)
= 46.(-86 -14)
= 46.(-100)
= -4600
c,
-72(15 - 49) + 15(-56 + 72)
= -72.15 + 72.49 - 56.15 + 72.15
= 56(63 - 15)
= 2688
d,
(-24).17.(-3)0.(-5)6.(-12n)
= (-24).17.(-56)
= 4250000
Đặt \(S=7+7^2+7^3+...+7^{52}\)
\(\Rightarrow7S=7^2+7^3+7^4+...+7^{53}\)
\(\Rightarrow7S-S=6S=7^2+7^3+7^4+...+7^{53}-\left(7+7^2+7^3+...+7^{52}\right)\)
\(\Leftrightarrow6S=7^{53}-7\)
\(\Rightarrow S=\frac{7^{53}-7}{6}\)
Thay vào biểu thức ta có:
\(\frac{6x\left(7^{53}-7\right)}{6}+7=7^{53}.\)
\(\Leftrightarrow x\left(7^{53}-7\right)=7^{53}-7\)
\(\Rightarrow x=1\)
đặt A=\(7+7^2+...+7^{52}\)
7A =\(7^2+7^3+...+7^{53}\)
=>7A-A=\(\left(7^2+7^3+...+7^{53}\right)-\left(7+7^2+...+7^{52}\right)\)
6A=\(7^{53}+7\)
A=\(\frac{7^{53}+7}{6}\)
TA CÓ : 6X.\(\left(7+7^2+...+7^{52}\right)\)+7=\(7^{53}\)
(=)6X.\(\frac{7^{53}+7}{6}\)+7=\(7^{53}\)
(=)6X.\(\frac{7^{53}+7}{6}\)=\(7^{53}-7\)
(=)X.\(7.\left(7^{52}+1\right)\)=\(7.\left(7^{52}-1\right)\)
X=7.\(\left(7^{52}-1\right):[7.\left(7^{52}+1]\right)\)
X=\(\frac{7^{52}-1}{7^{52}+1}\)
a.=18 : 9 + 5 .8
=2+40
=42
b. = -(12+42)
= -54
c. =53.25+53.75+27.18+27.82
=53.(25+75)+27.(18+82)
=53.100+27.100
=100.(53+27)
=100.80
=8000
d.{75+(-75)}+(-350)
=0+(-350)
=-350
a) 18 : 32 + 5 . 23
= 18 : 9 + 5 . 8
= 2 + 40
= 42
b) (-12) + (-42)
= -(12+42)
= - 54
c) 53.25 + 53.75 + 27.18 + 33.82
= 53.25 + 53.75 + 27 .18 + 27.82
= 53.(25+75) + 27.(18+82)
= 53.100 + 27.100
= 100.(53 + 27)
= 100. 80
= 8000
d) 75 + (-350) + (-75)
= 75 + (-75) + (-350)
= 0 + (-350)
= -350
Chúc bạn học tốt!
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)
2 . n3 - 1 = 53
2 . n3 = 53 + 1
2 . n3 = 54
n3 = 54 : 2
n3 = 27
n3 = 33
=> n = 3