\(x=\frac{\left(6+3^2.2-6.3^2\right)^2}{\left(3^2+3.3^2-3^4\right)^2}\)
\(x=\frac{15.3^{11}+4.2.7^4}{9^7}\)
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30 tháng 7 2018
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
30 tháng 3 2018
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
13 tháng 8 2015
a) \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)
B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)
BÀI 2:
A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)
\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)
\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)
\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)
b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)
\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)
ngoặc 1 có 99 số hạng x
\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3.\frac{99}{100}=1\)
\(\Leftrightarrow99x=1+\frac{3.99}{100}\)
\(\Leftrightarrow99x=\frac{397}{100}\)
\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)
\(x=\frac{\left[6+3^2.2-6.3^2\right]^2}{\left[3^2+3.3^2-3^4\right]^2}=\frac{\left[6\left[1+3-9\right]\right]^2}{\left[9+27-81\right]^2}=\frac{\left[-30\right]^2}{\left[-45\right]^2}=\frac{900}{2025}=\frac{4}{9}\)