(*Không chép lại A nhé, mỏi tay :V*)

\(A=\frac{21-18}{18\cdot21}+\frac{24-21}{21\cdot24}+...+\frac{90-87}{87\cdot90}\\ =\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\\ =\frac{1}{18}-\frac{1}{90}=\frac{2}{45}\)

Ez, chúc bạn học tốt nha.

Thanks

\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72\left(not27\right)}=\frac{5.18-10.27+15.36}{4\left(5.18-10.27+15.36\right)}=\frac{1}{4}\)

\(\frac{\frac{-6}{7}+\frac{6}{19}-\frac{6}{31}}{\frac{9}{7}-\frac{9}{19}+\frac{9}{31}}=\frac{-6\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}{9\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)

\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)

Thank you <3

Sử dụng khá nhiều kiến thức hằng đẳng thức lớp 8, lớp 7 bó tay

\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+...-\frac{197^3}{9702}+\frac{199^3}{9900}\)

\(\frac{A}{2}=\frac{3^3}{1.2}-\frac{5^3}{2.3}+\frac{7^3}{3.4}-\frac{9^3}{4.5}+...+\frac{199^3}{99.100}\)

\(\frac{A}{2}=3^3\left(1-\frac{1}{2}\right)-5^3\left(\frac{1}{2}-\frac{1}{3}\right)+7^3\left(\frac{1}{3}-\frac{1}{4}\right)-...+199^3\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{A}{2}=3^3-\frac{3^3+5^3}{2}+\frac{5^3+7^3}{3}-\frac{7^3+9^3}{4}+...+\frac{197^3+199^3}{99}-\frac{199^3}{100}\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}-\left(16.2^2+12\right)+\left(16.3^2+12\right)-\left(16.4^2+12\right)+...+\left(16.99^2+12\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(3^2-2^2+5^2-4^2+7^2-6^2+...+99^2-98^2\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(2+3+4+5+...+98+99\right)\)

\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(99.50-1\right)\)

\(\Rightarrow A=16.99.100-\frac{199^3}{50}+22\) (đến đây bấm máy ra kết quả so sánh cũng được)

\(\Rightarrow A=\frac{2^3.100^2\left(100-1\right)-199^3}{50}+22\)

\(A=\frac{200^3-199^3-2.200^2}{50}+22\)

\(A=\frac{200^2+200.199+199^2-2.200^2}{50}+22\)

\(A=\frac{199^2-200^2+200.199}{50}+22\)

\(A=\frac{-199-200+200.199}{50}+22=\frac{199^2}{50}+18\)

\(A< \frac{199.200}{50}+18=814\)

Vậy \(A< 814\)

b)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(2.0,4\right)^5}{\left(0,4\right)^6}=\frac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\frac{2^5}{0,4}=\frac{2^5}{\frac{2}{5}}=\frac{2^4}{5}=\frac{16}{5}\)

c)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^{12}}{3^6.2^6.2^9}=3^6\)

a)\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{15^{10}.3^{10}.5^{20}}{5^{15}.15^{15}}=\frac{3^{10}.5^5}{15^5}=\frac{3^{10}.5^5}{5^5.3^5}=3^5\)

Thanks bn!!

Xin lỗi nha mk ms lp 6 ak nên ko hỉu.

Bn trả lời câu hỏi bên dưới của mk ik mk k cho!

Ta có : \(\frac{1+2y}{18}=\frac{1+4y}{24}\)

\(\Rightarrow24.\left(1+2y\right)=18.\left(1+4y\right)\)

\(\Rightarrow24+48y=18+72y\)

\(\Rightarrow24-18=72y-48y\)

\(\Rightarrow6=24y\Rightarrow y=\frac{6}{24}=\frac{1}{4}\)

Thay y vào đẳng thức ta có:

\(\frac{1+4.\frac{1}{4}}{24}=\frac{1+6.\frac{1}{4}}{6x}\)

\(\Rightarrow\frac{1}{12}=\frac{5}{2}:6x\)

\(\Rightarrow6x=\frac{5}{2}:\frac{1}{12}=30\)

\(\Rightarrow x=30:6=5\)

\(a.=\frac{3}{15}+\frac{-10}{15}\)

\(=-\frac{7}{15}\)

\(b.=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=1+\left(-1\right)\)

\(=0\)

\(c.=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)

\(=-1+1-\frac{1}{2}\)

\(=0-\frac{1}{2}\)

\(=-\frac{1}{2}\)

\(d.=\frac{5}{6}.\left(18\frac{2}{3}-6\frac{2}{3}\right)\)

\(=\frac{5}{6}.12\)

\(=10\)