Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Khó thế
làm thôi chứ mình nghĩ quá với lớp 6
\(\left\{{}\begin{matrix}\dfrac{4}{7}< \dfrac{a}{b}< \dfrac{2}{3}\\7a+4b=1994\end{matrix}\right.\)
\(7a+4b=1994\Rightarrow a=\dfrac{1994-4b}{7}=\dfrac{1998-4-4b}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}b=7n-2\\a=286-4n\end{matrix}\right.\)(*)
\(\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{2}{3}\left(1\right)\\\dfrac{a}{b}>\dfrac{4}{7}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3a< 2b\Rightarrow3\left(286-4n\right)< 2\left(7n-2\right)\)
\(858-12n< 14n-4\Rightarrow n>\dfrac{862}{26}=33,15..\)
\(\left(2\right)\Leftrightarrow7a>4b\Rightarrow7\left(286-4n\right)>2\left(7n-2\right)\)
\(\Leftrightarrow1932-28n>14n-4\Rightarrow N< \dfrac{1936}{42}=46,09..\)
\(\)Tập hợp giá trị phân số a/b thỏa mãn là:
\(\left\{{}\begin{matrix}n\in N\\33< n\le46\\\dfrac{a}{b}=\dfrac{286-4n}{7n-2}\end{matrix}\right.\)
thích bao nhiêu thay vào
ví dụ
\(\left\{{}\begin{matrix}n=34\\\dfrac{a}{b}=\dfrac{286-4.34}{7.34-2}=\dfrac{150}{236}\\7.150+4.236=1050+944=1994\end{matrix}\right.\)
cảm ơn bn nhìu nhé!!!!!!!!ko có bn mk ko bít làm thế nào lun á!!!!!!!!!
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)
\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)
\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{10}\)
\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)
\(\Leftrightarrow D< 1\left(đpcm\right)\)
\(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\cdot\cdot\cdot\left(1+\dfrac{7}{180}\right)=\dfrac{16}{9}\cdot\dfrac{27}{20}\cdot\cdot\cdot\dfrac{187}{180}=\dfrac{2.8}{1\cdot9}\cdot\dfrac{3\cdot9}{2\cdot10}\cdot\cdot\cdot\dfrac{11\cdot17}{10\cdot18}=\dfrac{\left(2\cdot3\cdot...\cdot11\right)\cdot\left(8\cdot9\cdot...\cdot17\right)}{\left(1\cdot2\cdot...\cdot10\right)\cdot\left(9\cdot10\cdot...\cdot18\right)}=\dfrac{11\cdot8}{1\cdot18}=\dfrac{88}{18}=\dfrac{44}{9}\)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
= \(\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+1-\dfrac{1}{4}+...+1-\dfrac{1}{100}}\)
= \(\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
=\(\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)= 2
7a+4b=1994
7a=1994-4b
7a=997.2-2b-2b
7a=2.(997-2b)
=[2.(997-2b)] :7
=[2.(997-2b)] : (3+4)(1)
7a+4b=1994
4b=1994-7a
4b=2.997-2a-5a
4b=2.(997-2a)-5a
= [2.(997-2a)-5a]:4(2)
từ (1),(2)
4/7<[2.(997-2b)]:7/[2.(997-2a)-5a]:4<2/3
a: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)
a) Vì -11 < -10 < -9 < -8 < -7 nên:
.
b) Quy đồng mẫu các phân số ta có:
Vì -12 < -11 < -10 < -9 nên ta có:
hay
a) Vì -11 < -10 < -9 < -8 < -7 nên:
.
b) Quy đồng mẫu các phân số ta có:
Vì -12 < -11 < -10 < -9 nên ta có:
hay
thanh zai dc