Tìm x nguyên biết 4x+9/6x+5 nguyên
Tìm số tự nhiên n biết 1/3+1/6+1/10+...+2/n(n+1)=2003/2004
CMR:5/9<1/15+1/16+...+1/33+1/34<4/3
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ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
14 tháng 7 2023
b: Để A nguyên thì 2n+3 chia hết cho n
=>3 chia hết cho n
=>n thuộc {1;-1;3;-3}
c: Th1: n=2
=>n+3=5(nhận)
TH2: n=2k+1
=>n+3=2k+4=2(k+2)
=>Loại
d: Gọi d=ƯCLN(2n+3;2n+5)
=>2n+5-2n-3 chia hết cho d
=>2 chia hết cho d
mà 2n+3 lẻ
nên d=1
=>PSTG
NH
1
15 tháng 12 2017
1/3 + 1/6 + 1/10 + ... + 2/n(n+1) = 2003/2004
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)
\(\Rightarrow n+1=4008\)
\(\Rightarrow n=4008-1=4007\)
HM
1
27 tháng 1 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{2003}{2004}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Leftrightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}=\frac{1}{4008}\)
\(\Rightarrow n+1=4008\Rightarrow n=4007\)
Vậy \(n=4007\)
19 tháng 10 2021
Câu 1:
\(a,=43\cdot\left(27+93\right)+3111+3363=43\cdot120+6474=11634\\ b,=11^2+2^{15}\cdot2^3:2^{17}=121+2=123\\ c,=11^2+7^2-9=121+49-9=151\)
Câu 2:
\(a,\Rightarrow x-\dfrac{3}{2}=5^2=25\\ \Rightarrow x=25+\dfrac{3}{2}=\dfrac{53}{2}\\ b,\Rightarrow7x=30-2=28\\ \Rightarrow x=4\)
Để \(\frac{4x+9}{6x+5}\)\(\in Z\)thì \(4x+9\)chia hết \(6x+5\)
\(\Rightarrow3.\left(4x+9\right)\)chia hết cho \(6x+5\)
\(\Rightarrow\)\(12x+27\)chia hết cho \(6x+5\)
\(\Rightarrow\)\(2.\left(6x+5\right)+17\)chia hết cho \(6x+5\)
\(\Rightarrow\)17 chia hết cho \(6x+5\)
\(\Rightarrow\)6x +5 thuộc Ư(17)
suy ra 6x+5 thuộc {+-1;+-17}
ĐẾN ĐÂY BẠN TỰ LẬP BẲNG TÌM X NHÉ
Vậy x thuộc{-1;2}
B)Tích đi mình làm tiếp cho
Có: 1/3+1/6+1/10+...+2/n(n+1)=2003/2004
=>1/2.[ 1/3+1/6+1/10+...+2/n(n+1)]=2003/2004.1/2
=>1/6+1/12+1/20+...+1/n.(n+1)=2003/2004.1/2
=>1/2.3+1/3.4+1/4.5+...+1/n.(n+1)=2003/2004.1/2
=>1/2-1/3+1/3-1/4+1/4-1/5+....+1/n-1/n+1=2003/2004.1/2
=>1/2-1/n+1=2003/4008
=>1/n+1=1/4008
=>n+1=4008
=>n=4007
Vậy n=4007