Gọi số cần tìm là \(\overline{ab}\) theo đề bài

\(\overline{a0b}=7x\overline{ab}\)

\(\Rightarrow100xa+b=70xa+7xb\)

\(\Rightarrow30xa=6xb\Rightarrow5xa=b\Rightarrow a=1;b=5\)

Số cần tìm là 15

Bạn ghi đúng môn học nhé.

\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\left(\dfrac{3}{2}\right).\left(\dfrac{4}{3}\right).\left(\dfrac{5}{4}\right)...\left(\dfrac{100}{99}\right)\)

\(=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=50\)

Mình giải ở câu hỏi ban này rồi nhé

\(\overline{ab}+\overline{ac}+\overline{ba}+\overline{bc}+\overline{ca}+\overline{cb}=528\)

\(10xa+b+10xa+c+10xb+a+10xb+c+10xc+a+10xc+b=528\)

\(22x\left(a+b+c\right)=528\)

\(\Rightarrow a+b+c=24\)

\(\Rightarrow a=7;b=8;c=9\)

14

35:5x2=14

Ta có: \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=\dfrac{100}{2}\)

\(=50\)

Tìm x hả bạn?

Giải: Ta có: \(1+3+5+...+\left(2x-1\right)=225\)

\(\Leftrightarrow\dfrac{\left(1+\left(2x-1\right)\right)x}{2}=225\)

\(\Leftrightarrow\dfrac{2x.x}{2}=225\)

\(\Leftrightarrow\dfrac{2x^2}{2}=225\Leftrightarrow x^2=225\)

\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-15\end{matrix}\right.\)

Ta thấy x>0 => x=-15 không thoả mãn

Vậy x=15

Ta có: \(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\)

\(=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=1:\left(\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)=1:\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)