( 1/5 + x  )  mũ 2= 9/16

<=>(1/5+x)mũ 2=(3/4)mũ 2

=>1/5+x=3/4

<=>x=3/4-1/5

<=>x=11/20

vậy x=..............

\(\left(\frac{1}{5}+x\right)^2=\frac{9}{16}\)

<=>\(\left(\frac{1}{5}+x\right)^2-\left(\frac{3}{4}\right)^2=0\)

<=>\(\left(\frac{1}{5}+x-\frac{3}{4}\right)\cdot\left(\frac{1}{5}+x+\frac{3}{4}\right)=0\)

<=>\(\left(x-\frac{11}{20}\right)\cdot\left(x+\frac{19}{20}\right)=0\)

<=>\(\orbr{\begin{cases}x-\frac{11}{20}=0\\x+\frac{19}{20}=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{11}{20}\\x=-\frac{19}{20}\end{cases}}\)

Vậy ...............

* Hình tự vẽ

a) Theo đề ra: Hai đường tròn ( A; 3cm ) và ( B; 3cm ) cắt nhau tại hai điểm C, D nên ta có: AC = BC = BD = DA = 3cm

=> A thuộc đường tròn ( C; 3cm )

b) Theo phần a), ta có: AC = BC = BD = DA = 3cm

=> Đường tròn ( D; 3cm ) đi qua hai điểm A, B

CẢM ƠN BẠN NHÉ ! ^ v ^

\(S=\frac{1}{2022}-\frac{5}{2.4}-\frac{5}{4.6}-...-\frac{5}{2020.2022}\)

\(=\frac{1}{2022}-\frac{5}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2020.2022}\right)\)

\(=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)

\(=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=\frac{1}{2022}-\frac{5}{2}.\frac{1010}{2022}=\frac{1}{2022}-\frac{2525}{2022}=\frac{-1262}{1011}\)

S = 1/2022 - 5/2x4 - 5/4x6 - ... - 5/2020x2022

= 1/2022 - (5/2x4 + 5/4x6 + ... + 5/2020x2022)

= 1/2022 - (2/2x4 + 2/4x6 + ... + 2/2020x2022) x 5/2

= 1/2022 - (1/2 - 1/4 + 1/4 - 1/6 + ... + 1/2020 - 1/2022) x 5/2

= 1/2022 - (1/2 - 1/2022) x 5/2

= 1/2022 - 1010/2022 x 5/2

= 1/2022 - 2525/2022

= 2024/2022 = 1012/1011

Đặt A = \(\frac{10^{2018}+1}{10^{2019}+1}\)

=> 10A = \(\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)

Đặt B = \(\frac{10^{2019}+1}{10^{2020}+1}\)

=> 10B = \(\frac{10.\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)

Nhận thấy 10²⁰¹⁹ + 1 < 10²⁰²⁰ + 1

=> \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\)

=> \(1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)

=> 10A > 10B

=> A > B

Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)

= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)

= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)

= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)

= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)

An góp được 3/5 số tiền mua bóng . Suy ra Bình góp được 2/5 số tiền mua bóng:

a)Vậy quả bóng đó giá:

80 000 : 2/5 = 200 000 ( đồng)

b) Số tiền An góp là:

200 000 - 80 000 = 120 000 ( đồng)

Đ/s; a) 200 000 đồng

          b) 120 000 đồng

\(A=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+\frac{28}{\left(6.8\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(100.98\right)^2}\)

\(=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+\frac{8^2-6^2}{6^2.8^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{6^2}-\frac{1}{8^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(=\frac{1}{2^2}-\frac{1}{100^2}< \frac{1}{4}\)

A = \(\frac{12}{\left(2.4\right)^2}\) +  \(\frac{20}{\left(4.6\right)^2}\) + ..........+ \(\frac{388}{\left(96.98\right)^2}\) +  \(\frac{396}{\left(98.100\right)^2}\)

   = \(\frac{16-4}{\left(2.4\right)^2}\)+ \(\frac{36-16}{\left(4.6\right)^2}\)+...........+ \(\frac{9604-9216}{\left(96.98\right)^2}\) + \(\frac{10000-9604}{\left(98.100\right)^2}\)

   = \(\frac{1}{2^2}\) - \(\frac{1}{4^2}\)+  \(\frac{1}{4^2}\)- \(\frac{1}{6^2}\) + ............+ \(\frac{1}{96^2}\) - \(\frac{1}{98^2}\) + \(\frac{1}{98^2}\) - \(\frac{1}{100^2}\)

   = \(\frac{1}{2^2}\) - \(\frac{1}{100^2}\)

   = \(\frac{1}{4}\) - \(\frac{1}{100^2}\) < \(\frac{1}{4}\)

   =) A < \(\frac{1}{4}\)

Chúc bạn học tốt nhé !

Ta có: a chia cho 2 dư 1 => a - 1 ⋮2

         a chia cho 3 dư 1 => a - 1 ⋮3

=> a - 1 ⋮6 => a -1 + 6.2 ⋮ 6 => a +11 ⋮ 6 (1)

Ta có: a  chia 5 dư 4 => a - 4 ⋮5 => a - 4 + 5.3 ⋮5  => a + 11 ⋮5 (2)

Ta có: a chia 7 dư 3 => a - 3 ⋮7 => a - 3 + 7.2 ⋮7 => a + 11 ⋮7 (3)

Từ (1) ; (2) ; (3) => a +11 ∈∈BC ( 6; 5; 7 ) 

Có: BCNN ( 6; 5; 7 ) = 210 

=> a + 11 ∈ BC ( 6; 5; 7 ) 

=> a ∈ { 199; 409 ;....}

Mà a là số tự nhiên nhỏ nhất nên a = 199.

sai òi bn oi:<

ko cs chia cho 6

T^T