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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
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
7a+4b=1994 (1), chia cả 2 vế cho b => 7a/b+4=1994/b <=> 7a/b=(1994-4b)/b
<=> a/b=(1994-4b)/7b
Theo bài ra có: (1994-4b)/7b > 4/7 => 1994-4b>4b
<=> 1994> 8b => b < 1994:8=> b<249,25
Lại có: (1994-4b)/7b < 2/3 <=> 3(1994-4b)<14b
<=> 26b>5982 => b> 5982:26=230,07
=> b=(231; 232; 233; ....; 249)
Thay vào (1) => a, chọn a thuộc N
=> tìm đc a/b
7a+4b=1994 (1), chia cả 2 vế cho b => 7a/b+4=1994/b <=> 7a/b=(1994-4b)/b
<=> a/b=(1994-4b)/7b
Theo bài ra có: (1994-4b)/7b > 4/7 => 1994-4b>4b
<=> 1994> 8b => b < 1994:8=> b<249,25
Lại có: (1994-4b)/7b < 2/3 <=> 3(1994-4b)<14b
<=> 26b>5982 => b> 5982:26=230,07
=> b=(231; 232; 233; ....; 249)
Thay vào (1) => a, chọn a thuộc N
=> tìm đc a/b
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}< x< \dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}< x< \dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}\)
\(\Leftrightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(\Leftrightarrow-2< 24x< 3\)
=>x=0
b: \(\Leftrightarrow\dfrac{9-10}{12}< \dfrac{x}{12}< 1-\dfrac{8-3}{12}=\dfrac{7}{12}\)
=>-1<x<7
hay \(x\in\left\{0;1;2;3;4;5;6\right\}\)
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 \(x\in\left\{-2;-1;0;1;2;3\right\}\)
Ta có:
\(2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(=2.1+2.\dfrac{1}{2}+2.\dfrac{1}{4}+2.\dfrac{1}{8}+...+2.\dfrac{1}{1024}\)
\(=2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)
Ta lại có:
\(A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)
\(=2-\dfrac{1}{1024}=\dfrac{2047}{1024}\)
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
\(\left\{{}\begin{matrix}n=34\\\dfrac{a}{b}=\dfrac{286-4.34}{7.34-2}=\dfrac{150}{236}\\7.150+4.236=1050+944=1994\end{matrix}\right.\)
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