cho a,b là các số nguyên cmr nếu (2a+3b)⋮7 thì (8a+5b)⋮7
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>a\(\le\)3b
=>b\(\le\)4c
=>c\(\le\)5d
=>d\(\le\)2023
do đó a\(\le\)3.4.5.2023=121380
vì a lớn nhất nên a=121380
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{7}+\dfrac{1}{y}=-\dfrac{1}{14}\)
=>\(\dfrac{xy+7}{7y}=\dfrac{-1}{14}\)
=>\(\dfrac{xy+7}{y}=\dfrac{-1}{2}\)
=>\(2\left(xy+7\right)=-y\)
=>2xy+y=-14
=>y(2x+1)=-14
=>\(\left(2x+1;y\right)\in\left\{\left(1;-14\right);\left(-14;1\right);\left(-1;14\right);\left(14;-1\right);\left(2;-7\right);\left(-7;2\right);\left(-2;7\right);\left(7;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;-14\right);\left(-15;1\right);\left(-1;14\right);\left(\dfrac{13}{2};-1\right);\left(\dfrac{1}{2};-7\right);\left(-8;2\right);\left(-\dfrac{3}{2};7\right);\left(3;-2\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-\dfrac{2}{3}x+\dfrac{3}{5}=1\dfrac{1}{3}\\ -\dfrac{2}{3}x+\dfrac{3}{5}=\dfrac{4}{3}\\ -\dfrac{2}{3}x=\dfrac{11}{15}\\ x=-\dfrac{11}{10}\\ b,\dfrac{1}{3}:x-\dfrac{2}{3}=5\\\dfrac{1}{3}:x=\dfrac{17}{3}\\ x=\dfrac{17}{9}\\ c,\left(x+3,6\right):0,3=9,6\\ x+3,6=2,88\\ x=-0,72.\\ d,x+14,12-33,2=66,8\\ x-19,08=66,8\\ x=85,88 \)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a; \(\dfrac{-24}{11}\) + \(\dfrac{-19}{13}\) - (\(\dfrac{-2}{11}\) + \(\dfrac{20}{13}\))
= - \(\dfrac{24}{11}\) - \(\dfrac{19}{13}\) + \(\dfrac{2}{11}\) - \(\dfrac{20}{13}\)
= - (\(\dfrac{24}{11}\) - \(\dfrac{2}{11}\)) - (\(\dfrac{19}{13}\) + \(\dfrac{20}{13}\))
= - \(\dfrac{22}{11}\) - \(\dfrac{39}{13}\)
= - 2 - 3
= - 5
Bài 6
a; A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\) + ... + \(\dfrac{1}{50^2}\)
\(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)
\(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}-\dfrac{1}{4}\)
\(\dfrac{1}{5^2}\) < \(\dfrac{1}{4.5}\) = \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)
.....................................
\(\dfrac{1}{50^2}\) < \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)
Cộng vế với vế ta có:
A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\) < \(\dfrac{1}{9}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{50}\) = \(\dfrac{4}{9}\) - \(\dfrac{1}{50}\) < \(\dfrac{4}{9}\) (1)
\(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)
\(\dfrac{1}{4^2}\) > \(\dfrac{1}{4.5}\) = \(\dfrac{1}{4}-\dfrac{1}{5}\)
....................................
\(\dfrac{1}{50^2}\) > \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)
Cộng vế với vế ta có:
A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\) > \(\dfrac{1}{9}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{50}\) = \(\dfrac{1}{4}\) + (\(\dfrac{1}{9}\) - \(\dfrac{1}{50}\)) > \(\dfrac{1}{4}\) (2)
Kết hợp (1) và (2) ta có: \(\dfrac{1}{4}\) < A < \(\dfrac{4}{9}\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: Điểm B thuộc tia Oy
a: Các tia đối nhau gốc O là Ox,Oy; OA,OB; OA,Oy; OB,Ox
b: Các tia đối nhau gốc A là Ax,Ay
c: OA và OB là hai tia đối nhau
=>O nằm giữa A và B
=>OA+OB=AB
=>OB+3,4=7
=>OB=3,6(cm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(AE+EQ+QC=AC\Rightarrow\dfrac{1}{4}EQ+EQ+\dfrac{1}{2}EQ=AC\)
\(\Rightarrow\dfrac{7}{4}EQ=AC\Rightarrow EQ=\dfrac{4}{7}AC\)
Hai tam giác ACD và DEQ chung đỉnh D và đáy cùng nằm trên đường thẳng AC
\(\Rightarrow S_{DEQ}=\dfrac{4}{7}S_{ACD}\)
Hai tam giác ABC và BEQ có chung đỉnh B và đáy cùng nằm trên đường thẳng AC
\(\Rightarrow S_{BEQ}=\dfrac{4}{7}S_{ABC}\)
\(\Rightarrow S_{DEQ}+S_{BEQ}=\dfrac{4}{7}S_{ACD}+\dfrac{4}{7}S_{ABC}\)
\(\Rightarrow S_{EDQB}=\dfrac{4}{7}S_{ABCD}=\dfrac{4}{7}.812=464\left(cm^2\right)\)
\(\left(2a+3b\right)⋮7\Rightarrow4\left(2a+3b\right)⋮7\)
\(\Rightarrow\left(8a+12b\right)⋮7\)
\(\Rightarrow\left(8a+5b+7b\right)⋮7\)
Mà \(7b⋮7\Rightarrow\left(8a+5b\right)⋮7\)