Violympic toán 7

Lời giải:

Nếu số đó có 1 chữ số thì có thể là: $9$ 

Nếu số đó có 2 chữ số thì có thể là: $99$
Nếu số đó có 3 chữ số thì có thể là: $234, 243, 342, 324, 432, 423, 333, 999$

Nếu số đó có 4 chữ số thì số được tạo thành từ 2,3,4,9 luôn > 1000 (loại)

Vậy có tổng cộng 10 số thỏa mãn.

Đáp án C.

A = 1 + 4 + 16 + 64 + ... + 4096

= 2⁰ + 2² + 2⁴ + 2⁶ + ... + 2¹²

4A = 2² + 2⁴ + 2⁴ + 2⁸ + ... + 2¹² + 2¹⁴

3A = 4A - A

= (2² + 2⁴ + 2⁶ + 2⁸ + ... + 2¹² + 2¹⁴) - (1 + 2² + 2⁴ + 2⁶ + ... + 2¹²)

= 2¹⁴ - 1

A = (2¹⁴ - 1) : 3

= 5461

Chọn A

sao bài nó giống với ,.. trong .. v ta 🤔

`#3107.101107`

\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{97\cdot99}\right)-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{97}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{9603}-\dfrac{1}{2}\cdot\dfrac{96}{97}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{9603}-\dfrac{96}{97}\right)\\ =\dfrac{1}{2}\cdot\left(-\dfrac{9502}{9603}\right)\\ =-\dfrac{4751}{9603}\)

Vậy, `B = -4751/9603.`

\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)

Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)

\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)

\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)

\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)

\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)

\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5} +...+\dfrac{1}{95}-\dfrac{1}{97}\)

\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)

\(2C=\dfrac{96}{97}\)

\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)

Thay C vào ta được:

\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)

\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)

\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)

\(99B=-\dfrac{4751}{97}\)

\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)

\(\dfrac{1}{7}+\dfrac{1}{91}+...+\dfrac{1}{1147}\)

\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+...+\dfrac{1}{31\cdot37}\)

\(=\dfrac{1}{6}\cdot\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{31\cdot37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\left(1-\dfrac{1}{37}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{36}{37}\)

\(=\dfrac{6}{37}\)

Vậy ...

#\(Toru\)

\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)

\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{98}{303}\)

\(=\dfrac{49}{303}\)

Vậy \(S=\dfrac{49}{303}\)

#\(Toru\)

\(< =>\dfrac{30x^2}{60}+\dfrac{20y^2}{60}+\dfrac{15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\\ < =>30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\\ < =>18x^2+8y^2+3z^2=0\)

có \(\left\{{}\begin{matrix}x^2\ge0\\y^2\ge0\\z^2\ge0\end{matrix}\right.< =>\left\{{}\begin{matrix}18x^2\ge0\\8y^2\ge0\\3z^2\ge0\end{matrix}\right.\)

`=>18x^2+8y^2+3z^2>=0`

dấu ''='' xảy khi \(\left\{{}\begin{matrix}18x^2=0\\8y^2=0\\3z^2=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

á bài khó qué 🫤

\(\dfrac{x^2}{2}+\dfrac{y^2}{3}+\dfrac{z^2}{4}=\dfrac{x^2+y^2+z^2}{5}\\ \Rightarrow\dfrac{30x^2+20y^2+15z^2}{60}=\dfrac{12x^2+12y^2+12z^2}{60}\\ \Rightarrow18x^2+8y^2+3z^2=0\)

Do \(x^2\ge0\forall x;y^2\ge0\forall y;z^2\ge0\forall z\)

\(\Rightarrow18x^2+8y^2+3z^2\ge0\)

Dấu "=" xảy ra khi \(x=y=z=0\)

\(\dfrac{x^2}{2}\)+\(\dfrac{y^2}{3}\) +\(\dfrac{z^2}{4}\) =\(\dfrac{x^2+y^2+z^2}{5}\) 

⇒\(30x^2\)+\(20y^2\)+\(15z^2\)=\(12x^2\)+\(12y^2\)+\(12z^2\) 

⇒\(18x^2\)+\(8y^2\)+ \(3z^2\)=0

⇒\(18x^2\)≥0 \(8y^2\)≥0 \(3z^2\)≥0

Nên \(18x^2\) + \(8y^2\) + \(3z^2\) ≥0

Vậy \(18x^2\) + \(8y^2\) + \(3z^2\)= 0

Khi và chỉ khi:

\(18x^2\)= 0;\(8y^2\)= 0;\(3z^2\)= 0

Vậy x=y=z=0

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=-\dfrac{11}{12}\)

\(B=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\left(3+\dfrac{1}{8}\right):\dfrac{75}{26}\)

\(=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{1}{3}\cdot\dfrac{13}{4}=\dfrac{13}{12}\)

b: A<x<B

=>-11/12<x<13/12

=>\(x\in\left\{0;1\right\}\)

a) \(A=\dfrac{1,11+0,19-1,3\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{3}\right):2\)

\(A=\dfrac{\left(1,11+0,19\right)-2,6}{2,6}-\left(\dfrac{3}{6}+\dfrac{2}{6}\right):2\)

\(A=\dfrac{1,3-2,6}{2,6}-\dfrac{5}{6}:2\)

\(A=\dfrac{-1,3}{2,6}-\dfrac{5}{6}:2\)

\(A=\dfrac{-1}{2}-\dfrac{5}{12}\)

\(A=\dfrac{-6}{12}-\dfrac{5}{12}\)

\(A=-\dfrac{11}{12}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)

\(B=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{23+26+26}{26}\)

\(B=\left(3+\dfrac{7}{8}-\dfrac{2}{8}-\dfrac{4}{8}\right):\dfrac{75}{26}\)

\(B=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}\)

\(B=\left(\dfrac{24}{8}+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}\)

\(B=\dfrac{25}{8}\cdot\dfrac{26}{75}\)

\(B=\dfrac{26}{24}\)

\(B=\dfrac{13}{12}\)

b) Ta có:

\(A< x< B\)

\(\Rightarrow-\dfrac{11}{12}< x< \dfrac{13}{12}\)

\(\Rightarrow x\in\left\{0;1\right\}\)