Gọi số cần tìm là a

Ta có: 

\(\overline{a0}+a+\dfrac{\overline{a0}}{a}=175\\ \Leftrightarrow10a+a+\dfrac{10a+0}{a}=175\\ \Leftrightarrow11a+10=175\\ \Leftrightarrow11a=175-10=165\\ \Rightarrow a=165:11=15\)

Vậy số cần tìm là 15

Ta đặt: A=1.2+2.3+3.4+...+100.101

       =>3A=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3A=100.101.102

       =>3A=1030200

       =>A=343400

 Chúc bạn hok tôt :)!!!!!

\(\dfrac{x-5}{x+7}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\\x+7< 0\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x-5< 0\\x+7>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>5\\x< -7\end{matrix}\right.\)  hoặc \(\left\{{}\begin{matrix}x< 5\\x>-7\end{matrix}\right.\)

\(\Leftrightarrow-7< x< 5\)

|x- 1/2| +2/3 = | 1/3 -2,5|

|x-1/2| + 2/3 = |-13/6|

| x -1/2|  + 2/3 = 13/6

| x-1/2| = 13/6 - 2/3

|x-1/2| = 3/2

x - 1/2 = + - 3/2

x = 2; x = - 1

vậy x ϵ {-1; 2}

Ta có:

\(x^2+2x=y^2\\ \Leftrightarrow x^2+2x+1=y^2+1\\ \Leftrightarrow\left(x+1\right)^2-y^2=1\\ \Leftrightarrow\left(x+1-y\right)\left(x+1+y\right)=1=1.1\)

Vì x,y nguyên nên x+1-y và x+1+y nguyên. 

Do đó: \(\left\{{}\begin{matrix}x+1-y=1\\x+1+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-y=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Cặp số nguyên (x,y) thõa mãn bài toán là (0;0)

     ( 10 - \(\dfrac{5}{3}\) + \(\dfrac{4}{9}\)) - ( 7- \(\dfrac{8}{3}\)- \(\dfrac{4}{9}\))

=    10 - \(\dfrac{5}{3}\) + \(\dfrac{4}{9}\) - 7 + \(\dfrac{8}{3}\) + \(\dfrac{4}{9}\)

= ( 10 - 7) +   ( \(\dfrac{8}{3}\)- \(\dfrac{5}{3}\)) + ( \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\))

= 3 + \(\dfrac{3}{3}\) + \(\dfrac{8}{9}\)

= 4 + \(\dfrac{8}{9}\)

= \(\dfrac{36}{9}\) + \(\dfrac{8}{9}\)

= \(\dfrac{44}{9}\)

TH1: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

\(\Rightarrow\left(2x-3\right)^2=3-2x\)

\(\Leftrightarrow\left(2x-3\right)^2+2x-3=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3+1\right)=0\)

Giải nhanh: \(x=\dfrac{3}{2}\)(nhận) hay \(x=1\) (nhận)

TH2: \(3-2x< 0\Leftrightarrow x>\dfrac{3}{2}\)

\(\Rightarrow\left(2x-3\right)^2=2x-3\)

Giải tương tự, ta suy ra \(x=\dfrac{3}{2}\) (loại) hay \(x=2\) (nhận)

Kết luận phương trình có 3 nghiệm \(x=1;x=\dfrac{3}{2};x=2\)

\(\dfrac{3}{5}.\dfrac{5}{7}+\dfrac{3}{5}:\dfrac{5}{3}-\dfrac{1}{7}:\dfrac{5}{3}\)

= \(\dfrac{3}{5}.\dfrac{5}{7}-\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{1}{7}.\dfrac{3}{5}\)

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

=\(\dfrac{3}{5}.1\)

=\(\dfrac{3}{5}\)