4/5.9/7-4/5.3/14
dấu chấm là nhân
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4/5.9/7-4/5.3/14
= 4/5 . (9/7 - 3/14)
= 4/5 . 15/14
= 6/7
Đề còn thiếu 1 điều kiện nữa là \(n>0\)
Đặt \(A=\frac{4}{5.2!}+\frac{4}{5.3!}+\frac{4}{5.4!}+...+\frac{4}{5.n!}\) ta có :
\(A=\frac{4}{5}\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\right)\)
Để \(A< 0,8\) thì \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}< 1\)
Đặt \(B=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\) ta có :
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}+\frac{1}{n}\)
\(B< 1-\frac{1}{n}< 1\)
\(\Rightarrow\)\(B< 1\) ( đpcm )
Suy ra : \(A=\frac{4}{5}.B=0,8.B< 0,8\) ( vì \(B< 1\) )
Vậy \(\frac{4}{5.2!}+\frac{4}{5.3!}+\frac{4}{5.4!}+...+\frac{4}{5.n!}< 0,8\)
Chúc bạn học tốt ~
( 4^5.9^4+2.6^9) : (2^10.3^8-6^8.2) = \(\frac{4^5.9^4+2.6^9}{2^{10}.3^8-6^8.2}=\frac{\left(2^2\right)^5.\left(3^2\right)^4+2.6^9}{2^{10}.3^8-6^8.2}\)
= \(\frac{2^{10}.3^8+2.6^9}{2^{10}.3^8-6^8.2}=\frac{2\left(6^8.8\right)}{2.6^8}=\frac{6^8.8}{6^8}=8\)
a) 5.3²⁰²³ = 50.3²⁰²³ - 5.9ˣ
5.9ˣ = 50.3²⁰²³ - 5.3²⁰²³
5.(3²)ˣ = 5.3²⁰²³.(10 - 1)
5.(3²)ˣ = 5.3²⁰²³.9
3²ˣ = 3²⁰²³.3²
3²ˣ = 3²⁰²⁵
2x = 2025
x = 2025/2
b) 2.3ˣ + 5.3ˣ⁺¹ = 153
3ˣ.(2 + 5.3) = 153
3ˣ.17 = 153
3ˣ = 153/17
3ˣ = 9
3ˣ = 3²
x = 2
Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)
\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)
x + 4 = 113 ⇒ x = 109
\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)
Với \(A=\dfrac{56}{113}\) thì
\(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)
\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=\dfrac{1}{113}\)
\(x=109\)
\(H=\frac{2\cdot2}{1\cdot5}+\frac{2\cdot2}{5\cdot9}+...+\frac{2\cdot2}{45.49}\)
\(H=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{45\cdot49}\)
\(H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(H=1-\frac{1}{49}\)
\(H=\frac{48}{49}\)
\(H=\frac{2.2}{1.5}+\frac{2.2}{5.9}+\frac{2.2}{9.13}+...+\frac{2.2}{45.49}\)
\(\Rightarrow H=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{45.49}\)
\(\Rightarrow H=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{49-45}{45.49}\)
\(\Rightarrow H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(\Rightarrow H=1-\frac{1}{49}=\frac{48}{49}\)
\(\left(x-4\right)^9=49.\left(x-4\right)^7\\ =>\left(x-4\right)^9:\left(x-4\right)^7=49\\ =>\left(x-4\right)^2=49\\ =>\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
(x - 4)9 = 49 . (x - 4)7
(x - 4)9 : (x - 4)7 = 49
(x - 4)2 = 72 = (-7)2
TH1 : TH2 :
(x - 4)2 = 72 (x - 4)2 = (-7)2
x - 4 = 7 x - 4 = -7
x = 7 + 4 x = -7 + 4
x = 11 x = -3
Vậy x = 11 Vậy x = -3
P = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{98}{99}\). CMR: P \(< \frac{1}{7}\)
Đề bài đây à
Viết phân số thích hợp vào chỗ chấm:
A) 3/4 nhân ( 5/6+6/5 )= 3/4 nhân 5/6+3/4 nhân 6/5
B) 7/5 nhân ( 4/7-2/5 )= 7/5 nhân 4/7- 7/5 nhân 2/5
4/5.9/7-4/5.3/14
= 4/5 . (9/7 - 3/14)
= 4/5 . 15/14
= 6/7
\(\frac{4}{5}\cdot\frac{9}{7}-\frac{4}{5}\cdot\frac{3}{14}=\frac{4}{5}\cdot(\frac{9}{7}-\frac{3}{14})=\frac{4}{5}\cdot\frac{15}{14}=\frac{6}{7}\)
P/S : Hoq chắc :>