\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{13}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{40}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{121}{81}+\dfrac{1}{243}+\dfrac{1}{720}\)
\(=\dfrac{364}{243}+\dfrac{1}{720}\)
\(=\dfrac{1092}{720}=\dfrac{91}{60}\)
hoặc có thể viết dưới dạng thập phân gần bằng \(1,5167\)

Bài 1:
Ta có \(2+4+6+...+\left(2n\right)=756\)
\(\Rightarrow\left(2n+2\right)n=756\)
\(\Leftrightarrow2n^2+2n-756=0\)
\(\Leftrightarrow\left(n-\dfrac{\sqrt{1513}-1}{2}\right)\left(n+\dfrac{\sqrt{1513}+1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n-\dfrac{\sqrt{1513}-1}{2}=0\\n+\dfrac{\sqrt{1513}+1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=\dfrac{\sqrt{1513}-1}{2}\\n=-\dfrac{\sqrt{1513}+1}{2}\end{matrix}\right.\)
Vậy \(n=\dfrac{\sqrt{1513}-1}{2}\) hoặc \(n=-\dfrac{\sqrt{1513}+1}{2}\)
Bài 2:
Ta có \(p=\left(n-2\right)\left(n^2+n-5\right)\)
Để \(p\) là số nguyên tố thì \(\left[{}\begin{matrix}n-2=1\\n^2+n-5=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3\\\left(n-2\right)\left(n+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3\\n-2=0\\n+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3\left(Nhận\right)\\n=2\left(Nhận\right)\\n=-3\left(Loại\right)\end{matrix}\right.\)
Vậy \(n=3\) hoặc \(n=2\)
 

a) \(\left|a+b\right|\le\left|a\right|+\left|b\right|\)
b) \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\) với \(\left|a\right|\ge\left|b\right|\)
c) \(\left|ab\right|\le\left|a\right|.\left|b\right|\)
d) \(\left|\dfrac{a}{b}\right|\le\dfrac{\left|a\right|}{\left|b\right|}\)

1/\(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(-20x+5\right)\)
\(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2+20x-5\)\(A=3\left(x^2+2x\left(-1\right)+\left(-1\right)^2\right)-\left(x^2+2x+1\right)+\left(2x-2.3\right)\left(x+3\right)-\left(\left(2x\right)^2+2.2x.3+3^2\right)+20x-5\)\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+\left(2x-6\right)\left(x+3\right)-\left(2^2.x^2+2.2.3x+9\right)+20x-5\)\(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+\left(2x-6\right)\left(x+3\right)-\left(4x^2+12x+9\right)+20x-5\)
\(A=3\left(x^2-2x+1\right)-x^2-2x-1+\left(2x-6\right)\left(x+3\right)-4x^2-12x-9+20-5\)
\(A=3\left(x^2-2x+1\right)-x^2-4x^2-2x-12x+20x+\left(2x-6\right)\left(x+3\right)-1-9-5\)
\(A=3\left(x^2-2x+1\right)-5x^2+6x+\left(2x-6\right)\left(x+3\right)-15\)
\(A=\left(3x^2-3.2x+3\right)-5x^2+6x+\left(2xx+2x.3-6x-6.3\right)-15\)
\(A=\left(3x^2-6x+3\right)-5x^2+6x+\left(2x^{1+1}+2.3x-6x-18\right)-15\)
\(A=\left(3x^2-6x+3\right)-5x^2+6x+\left(2x^2-18\right)-15\)
\(A=3x^2-6x+3-5x^2+6x+2x^2-18-15\)
\(A=3x^2-5x^2-2x^2-6x+6x+3-18-15\)
\(A=-30\)

= (3x - 2y)² + 2(3x - 2y)2 + 2²
= (3x - 2y + 2)²

\(x\in\left\{\dfrac{68}{9};\dfrac{23}{3};...;\dfrac{17}{2}\right\}\)

Ta có: 5+ 5² + 5³ + ... + 5¹²
= 1.(5+5²)+5²(5+5²)+ ... +5¹⁰(5+5²)

=1.30 + 5².30+ ... +5¹⁰.30

=30.(1 + 5² + ... + 5¹⁰) ⋮ 30

Vậy 5 + 5² + 5³ + ... + 5¹² ⋮ 30

Đặt a là chiều dài ; b là chiều rộng
ta được: \(\left\{{}\begin{matrix}b=\dfrac{a}{2}\\b=a-20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-20=\dfrac{1}{2}a\\b=a-20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a-40=a\\b=a-20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=20\end{matrix}\right.\)

Chu vi mảnh đất là:

(40+20)x2=120(m)

Diện tích mảnh đất là:

40x20=800(m²)

Vậy: Chu vi hình mảnh đất là 120 m

         Diện tích mảnh đất là 800 m²

Đặt a là số tuổi của em; b là số tuổi của anh
ta được: \(\left\{{}\begin{matrix}a+6=b\\5a=3b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b-6\\5\left(b-6\right)=3b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b-6\\2b=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=9\\b=15\end{matrix}\right.\)

Vậy số tuổi của em là 9, số tuổi của anh là 15