Gọi 2 số cần tìm là \(\overline{5ab};\overline{3cd}\) theo đề bài

\(\overline{5ab}+\overline{3cd}=836\)

\(500+\overline{ab}+300+\overline{cd}=836\)

\(\Rightarrow\overline{ab}+\overline{cd}=36\)

Đến đây là dạng toán tìm 2 số khi biết tổng và tỷ ở lớp 5

Chia 2 trường hợp \(\overline{ab}=2x\overline{cd}\) hoặc \(\overline{cd}=2x\overline{ab}\)

Bạn tự làm nốt nhé

\(\left(-x-\dfrac{1}{2}y\right)\left(-x+\dfrac{1}{2}y\right)\\ =\left[\left(-x\right)-\dfrac{1}{2}y\right]\left[\left(-x\right)+\dfrac{1}{2}y\right]\\ =\left(-x\right)^2-\left(\dfrac{1}{2}y\right)^2\\ =x^2-\dfrac{y^2}{4}\)

7\(x\).2 hay 7\(x^2\) vậy em?

7x^2 ạ

Ta có:

∠A₃ = ∠B₁ (gt)

Mà ∠A₃ và ∠B₁ là hai góc so le trong

⇒ a // b

⇒ ∠A₂ = ∠B₂ (đồng vị)

Do ∠A₃ + ∠A₄ = 180⁰ (kề bù)

Mà ∠A₃ = ∠B₁ (gt)

⇒ ∠B₁ + ∠A₄ = 180⁰

\(M=\dfrac{202}{7\cdot1010}+\dfrac{202}{10\cdot1313}+\dfrac{202}{13\cdot1616}+...+\dfrac{202}{91\cdot9494}\\ =\dfrac{202}{7\cdot10\cdot101}+\dfrac{202}{10\cdot13\cdot101}+\dfrac{202}{13\cdot16\cdot101}+...+\dfrac{202}{91\cdot94\cdot101}\\ =\dfrac{202}{101}\cdot\left(\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}+...+\dfrac{1}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\dfrac{87}{658}\\ =\dfrac{29}{329}\)

Ta có:

\(\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{15}\right|+...+\left|x-\dfrac{1}{399}\right|\ge0\forall x\)

\(\Rightarrow-11x\ge0\forall x\Rightarrow x\le0\)

\(\Rightarrow x-\dfrac{1}{3};x-\dfrac{1}{15};...;x-\dfrac{1}{399}< 0\)

\(\Rightarrow x-\dfrac{1}{3}+x-\dfrac{1}{15}+...+x-\dfrac{1}{399}=11x\)

\(\Rightarrow x+x+...+x-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right)=11x\)

Vì số lượng \(x\) ở vế trái bằng số lượng số hạng là phân số

\(\Rightarrow\) Số lượng \(x\) ở vế trái là:\(\left(19-1\right):2+1=10\left(số\right)\)

\(\Rightarrow10x-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=11x\)

\(\Rightarrow-x=1-\dfrac{1}{21}\)

\(\Rightarrow x=-\dfrac{20}{21}\)

\(a.\dfrac{-3}{7}\cdot\dfrac{15}{13}-\dfrac{3}{7}\cdot\dfrac{11}{13}-\dfrac{3}{7}\\ =-\dfrac{3}{7}\cdot\left(\dfrac{15}{13}+\dfrac{11}{13}+1\right)\\ =-\dfrac{3}{7}\cdot\left(\dfrac{26}{13}+1\right)\\ =\dfrac{-3}{7}\cdot3\\ =\dfrac{-9}{7}\\ b.\dfrac{-1}{9}\cdot\dfrac{-3}{5}+\dfrac{5}{-6}\cdot\dfrac{-3}{5}-\dfrac{7}{2}\cdot\dfrac{3}{5}\\ =-\dfrac{3}{5}\cdot\left(\dfrac{-1}{9}+\dfrac{-5}{6}+\dfrac{7}{2}\right)\\ =-\dfrac{3}{5}\cdot\dfrac{23}{9}\\ =-\dfrac{23}{15}\)

\(Ư\left(240\right)=\left\{\text{1;2,3,4,5,6,8,10,24,30,40,48,60,80,120,240}\right\}\)

240 = 2⁴.3.5 

Ư(240) ={1;2;3;4;5;6;8;10;12;15;16;20;24;30;40;48;60;80;120;240}

\(\overline{ab}+\overline{cd}+\overline{eg}⋮9\)

\(\overline{ab}+\overline{cd}+\overline{eg}=10a+b+10c+d+10e+g=\)

\(=9\left(a+c+e\right)+\left(a+b+c+d+e+g\right)⋮9\)

Ta có \(9\left(a+c+e\right)⋮9\)

\(\Rightarrow a+b+c+d+e+g⋮9\)

\(\Rightarrow\overline{abcdeg}⋮9\)

\(\dfrac{2}{1\cdot6}+\dfrac{2}{11\cdot16}+...+\dfrac{2}{x\left(x+5\right)}=\dfrac{41}{103}\\ =>\dfrac{2}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}\right)=\dfrac{41}{103}\\ =>1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{41}{103}:\dfrac{2}{5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=\dfrac{205}{206}\\ =>1-\dfrac{1}{x+5}=1-\dfrac{1}{206}\\ =>\dfrac{1}{x+5}=\dfrac{1}{206}\\ =>x+5=206\\ =>x=206-5=201\)