\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}=...\)
Những câu hỏi liên quan
\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)\cdot\cdot\cdot\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)\cdot\cdot\cdot\left(99\times102+2\right)}=...\)
\(\dfrac{\left(4\times7+2\right)\left(6\times6+2\right)\left(8\times11+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)\left(9\times12+2\right)...\left(99\times102+2\right)}\)
Mấy bạn giúp mình được không :((
\(\dfrac{4^{10}\times9^6+3^{12}\times8^5}{6^{13}\times4-2^{16}\times3^{12}}\)
\(\dfrac{2^4\times2^6}{\left(2^5\right)^2}-\dfrac{2^5\times15^3}{6^3\times10^2}\)
\(\dfrac{\left(-2\right)^{10}\times3^{31}+2^{40}\times\left(-3\right)^6}{\left(-2\right)^{11}\times\left(-3\right)^{31}+2^{41}\times3^6}\)
giải chi tiết giúp mình nhé
Tính\(\frac{\left(4\cdot7+2\right)\left(6\cdot9+2\right)\left(8\cdot11+2\right)..........\left(100\cdot103+2\right)}{\left(5\cdot8+2\right)\left(7\cdot10+2\right)\left(9\cdot12+2\right)..........\left(99\cdot102+2\right)}\)
Tình A=\(\frac{\left(4\cdot7+2\right)\left(6\cdot9+2\right)\left(8\cdot11+2\right)...\left(100\cdot103\right)}{\left(5\cdot8+2\right)\left(7\cdot10+2\right)\left(9\cdot12+2\right)...\left(99\cdot102+2\right)}\)
So sánh A và B , biết rằng:Asqrt[13]{12timesleft(11-frac{frac{left(10times8right)^{left(9+8right)}}{8}+frac{6^7times6-6}{6}}{frac{5}{4}}right)div3+2}Bsqrt[13]{12divleft(11+frac{frac{left(10times8right)^{left(9-8right)}}{8}+frac{6^7times6+6}{6}}{frac{5}{4}}right)times3-2}
Đọc tiếp
So sánh \(A\) và \(B\) , biết rằng:
\(A=\sqrt[13]{12\times\left(11-\frac{\frac{\left(10\times8\right)^{\left(9+8\right)}}{8}+\frac{6^7\times6-6}{6}}{\frac{5}{4}}\right)\div3+2}\)
\(B=\sqrt[13]{12\div\left(11+\frac{\frac{\left(10\times8\right)^{\left(9-8\right)}}{8}+\frac{6^7\times6+6}{6}}{\frac{5}{4}}\right)\times3-2}\)
frac{left(-7right)^n}{left(-7right)^{n-1}}(nge1) Tính GTBT Bài 2 Tính GTBT theo cách hợp lí nếu có thểc) frac{5^3times3^3}{5^3times0,5+125times2,5}d)frac{5times7^1+7^3times25}{7^5125-7^3times50}e)frac{8^5timesleft(-5right)^8+left(-2right)^5times10^9}{2^{16}times5^7+20^8}h)frac{left(-0,25right)^{-5}times9^4timesleft(-2right)^{-3}-2^{-2}times6^3}{2^9times3^6+6^6times40}Bài 3 Chứng tỏ rằnga)
Đọc tiếp
\(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)(n\(\ge1\)) Tính GTBT
Bài 2 Tính GTBT theo cách hợp lí nếu có thể
c) \(\frac{5^3\times3^3}{5^3\times0,5+125\times2,5}\)d)\(\frac{5\times7^1+7^3\times25}{7^5125-7^3\times50}\)e)\(\frac{8^5\times\left(-5\right)^8+\left(-2\right)^5\times10^9}{2^{16}\times5^7+20^8}\)
h)\(\frac{\left(-0,25\right)^{-5}\times9^4\times\left(-2\right)^{-3}-2^{-2}\times6^3}{2^9\times3^6+6^6\times40}\)
Bài 3 Chứng tỏ rằng
a)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{\left(3n-1\right)\times\left(3n+2\right)}=\frac{1}{6n+4}\)
b.\(\frac{5}{3\times7}+\frac{5}{7\times11}+\frac{5}{11\times15}+...+\frac{5}{\left(4n-1\right)\times\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
Đúng 0
Bình luận (0)
b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)
Đúng 0
Bình luận (0)