a: \(\frac{\left(-5\right)^{60}\cdot30^5}{15^5\cdot5^{61}}\)

\(=\frac{5^{60}}{5^{61}}\cdot\left(\frac{30}{15}\right)^5\)

\(=\frac{2^5}{5}=\frac{32}{5}\)

b: \(\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot\left(-1\right)\cdot3^{14}}=\frac{-3^{15}}{3^{14}}\cdot\frac15=-\frac35\)

c: \(4^6\cdot9^5+6^9\cdot120\)

\(=2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5\)

\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)

\(8^4\cdot3^{12}-6^{11}\)

\(=2^{12}\cdot3^{12}-2^{11}\cdot3^{11}\)

\(=2^{11}\cdot3^{11}\left(2\cdot3-1\right)=2^{11}\cdot3^{11}\cdot5\)

Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac45\)

d: \(4^2\cdot25^2+32\cdot125\)

\(=2^4\cdot5^4+2^5\cdot5^3\)

\(=2^4\cdot5^3\left(5+2\right)=2^4\cdot5^3\cdot7\)

\(\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}=2\cdot5\cdot7=10\cdot7=70\)

Ta có công thức tổng quát sau:

\(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)

\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có:\(\frac23\cdot\frac56\cdot\frac{9}{10}\cdot\ldots\cdot\frac{4949}{4950}\)

= \(\left(1-\frac13\right)\left(1-\frac16\right)\cdot\ldots\cdot\left(1-\frac{1}{4950}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\cdot\ldots\cdot\left(1-\frac{2}{9900}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot...\cdot\left(1-\frac{2}{99\cdot100}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(99+2\right)\left(99-1\right)}{99\left(99+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{101\cdot98}{99\cdot100}\)

\(=\frac{4\cdot5\cdot\ldots\cdot101}{3\cdot4\cdot\ldots\cdot100}\cdot\frac{1\cdot2\cdot\ldots\cdot98}{2\cdot3\cdot\ldots\cdot99}=\frac{101}{3}\cdot\frac{1}{99}=\frac{101}{297}\)

Tìm x

190−(x−1)2=189

(x - 1) ² = 190 - 189

(x - 1) ² = 1

x = 1+2

x = 2

a, cái câu trả lời trước mk ghi sai

190 - (x-1)² = 189

(x - 1 ) ² = 190 - 189

(x-1)² = 1

x = 1+1

x = 2

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

189 < 190 < 110

Đáp án cần chọn là B

B)TK:

+)(23.36-17.36):36

=[36.(23-17)]:36

=36.6:36

=(36:36).6
=1.6
=6

b: =23-17=6

Bạn ơi bạn ghi sai đề, đề đúng phải là:

C=198+194+190+...+6-196-192-188-...-8

=(6+...+190+194+198)-(8+...+188+192+196)

={(198+6)x[(198-6):4+1]:2}-{(196+8)x[(196-8):4+1]:2}

={204x49:2}-{204x48:2}

=204x49:2-204x48:2

=204x(49:2-48:2)

=204x0,5

=102

Chúc bạn học giỏi nha!!!

K cho mik vs nhé 

Ai lam cho mik nhah nhat mik k cho

\(A=\dfrac{4^2.25^2+32.125}{2^3.5^2}\)

\(A=\dfrac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{2^3.5^2}\)

\(A=\dfrac{2^4.5^4+2^5.5^3}{2^3.5^2}\)

\(A=\dfrac{2^4.2^5+5^4.5^3}{2^3.5^2}\)

\(A=\dfrac{2^{4+5}+5^{4+3}}{2^3.5^2}\)

\(A=\dfrac{2^9.5^7}{2^3.5^2}=2^6.5^5\)