Câu 1 :\(\frac{9.11+19.9}{27.5-27.15}\)
Câu 2 : \(\frac{6^5.8^2.21}{49}\)
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\(A=\frac{49^2\cdot3^{11}}{81^2\cdot21^5}\)
\(=\frac{\left(7^2\right)^2\cdot3^{11}}{\left(3^4\right)^2\cdot\left(3\cdot7\right)^5}\)
\(=\frac{7^4\cdot3^{11}}{3^8\cdot3^5\cdot7^5}\)
\(=\frac{7^4\cdot3^{11}}{3^{13}\cdot7^5}\)
\(=\frac{1}{3^2\cdot7}=\frac{1}{63}\)
Bài làm :
Ta có :
\(A=\frac{49^2\cdot3^{11}}{81^2\cdot21^5}\)
\(A=\frac{\left(7^2\right)^2\cdot3^{11}}{\left(3^4\right)^2\cdot\left(3\cdot7\right)^5}\)
\(A=\frac{7^4\cdot3^{11}}{3^8\cdot3^5\cdot7^5}\)
\(A=\frac{7^4\cdot3^{11}}{3^{13}\cdot7^5}\)
\(A=\frac{1}{3^2\cdot7}\)
\(A=\frac{1}{63}\)
Vậy A=1/63
a) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
Có:495.810/147.49.413=710.230/233.79=7/8 7/10-7/12+7/5 / 8/10-8/12+8/5=7(1/10-1/12+1/5) / 8(1/10-1/12+1/5)=7/8 =>A=7/8-7/8=0 xin lỗi nha mik ghi hơi khó hiểu chút vì mik mới dùng online math.HOK TỐT
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
\(\frac{9.11+19.9}{27.5-27.15}=\frac{9.\left(11+19\right)}{27.\left(5-15\right)}=\frac{9.30}{3.9.\left(-10\right)}=\frac{30}{3.\left(-10\right)}=\frac{30}{-30}=-1\)
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