Find n such that: 2/3 + 2/15 + 2/35 + ...... + 2/n = 322/323
Find n such that: 2/3+2/15+2/35+...+ 2/ nx(n+2)=322/323
2/3+2/15+2/35+...+2/n x (n+2)=322/323
=2/1.3+2.3.5+2/5.7+...+2/n x(n+2)=322/323
= n+2 =323
n=323-2
n=321
2/3+2/15+2/35+...+2/n x (n+2) = 322/323
2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323
1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323
1-1/n+2 = 322/323
1/n+2 = 1-322/323
1/n+2 = 1/323
=> n+2 = 323
n = 323 - 2 = 321
2/3 + 2/15+ 2/35+...+2/n = 322/323 tim n
2/3+2/15+2/35+...+2/n x (n+2) = 322/323
2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323
1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323
1-1/n+2 = 322/323
1/n+2 = 1-322/323
1/n+2 = 1/323
=> n+2 = 323
n = 323 - 2 = 321
tìm n biết
2/3+2/15+2/35+..........+2/n=322/323
Dấu / là gạch phân số.
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
Find the number n such that: 58-2*3^n =4 . Answer:
58-2*3^n+4<=>2*3^n=54<=>3^n=27<=>n=3
hence:n=3
Find n such that A= n^3-2n^2+2n-4 is a prime number?
In the figure below, x = Câu 2 Find the measure of angle J if \widehat{K}=29^o K =29 o and angle J and K are supplementary. Answer: \widehat{J}= J = ^o o . Câu 3 In the figure below, x = Câu 4 In the figure below, x = Câu 5 Solve: |2x-3|-4=3∣2x−3∣−4=3, where xx is a negative number. Answer: x=x= Câu 6 Calculate: 9+|-6|:1^2 =9+∣−6∣:1 2 = Câu 7 Compare: \dfrac{36}{27} 27 36 \dfrac{4}{3} 3 4 Câu 8 Find the negative number yy such that |y+\dfrac{7}{2}|-\dfrac{1}{2}=4∣y+ 2 7 ∣− 2 1 =4. Answer: y=y= Câu 9 Find xx such that \dfrac{32}{2^x}=2 2 x 32 =2. Answer: x=x= Câu 10 Find nn such that \dfrac{(-4)^n}{64}=-16 64 (−4) n =−16. Answer: n=n= Cần gấp nhé
find the value of n such that the polynomial 2x^5 - 3x^3 + x^2 +n is divisible by the polynomial x+2
tìm giá trị của n sao cho đa thức 2x ^ 5 - 3x ^ 3 + x ^ 2 + n chia hết cho đa thức x + 2
Tìm N biết
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{N}=\frac{332}{323}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{1}{1}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+2}{x+2}-\frac{1}{x+2}=\frac{332}{323}\)
=>\(\frac{x+1}{x+2}=\frac{332}{323}\)
=>332.(x+2)=323.(x+1)
=>332x+664=323x+323
=>332x-323x=323-664
=>x.(332-323)=-323
=>9x=-323
=>x=-323/9
vậy n=-323/9 .(-323/9+2)=98515/81
Find the sum A+B such that \(\frac{1}{n\left(n+2\right)}=\frac{A}{n+2}+\frac{B}{n}\)
Theo bài ra , ta có :
\(\frac{1}{n\left(n+2\right)}=\frac{A}{n+2}+\frac{B}{n}\)\(\left(ĐKXĐ:n\ne0;n\ne-2\right)\)
Quy đồng và khử mẫy ta được
\(An+B\left(n+2\right)=1\)
\(\Leftrightarrow An+Bn+2B=1\)
\(\Leftrightarrow n\left(A+B\right)+2B=1\)
\(\Leftrightarrow\hept{\begin{cases}n=1\\A+B=1\end{cases}}\)
Vậy A+B = 1
Chúc bạn học tốt =))
zZz Phan Cả Phát zZz tại s lại <=> n =1 và A+B= 1 v