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\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\left[33.\frac{4}{21}\right]\)
\(\Leftrightarrow A=\frac{7}{4}.\frac{44}{7}\)
\(\Leftrightarrow A=11\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)..................\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.............\frac{19}{20}\)
=\(\frac{1.2.3..............19}{2.3.4..............20}\)
=\(\frac{1}{20}\)
3333^4 =1111^4 x 3^4 =1111^4 x81
4444^3 =1111^3 x4^3 =1111^3 x64
=> 3333^4 >4444^3
\(A=\left(\frac{1-\left(\sqrt{a}\right)^3}{1-\sqrt{a}}\right)\left(\frac{1-\sqrt{a}}{1-\left(\sqrt{a}\right)^2}\right)^2\)
\(=\left(1+\sqrt{a}+a\right).\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\frac{1+\sqrt{a}+a}{1+2\sqrt{a}+a}\)
ta có:
\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)
\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)
\(=3log^2_2a+3log^a_2+1\)
=> A= \(\frac{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}\right).23.7.1009}{\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right).23.7.1009}\) + \(\frac{1}{30.1009-160}\)
=> A= \(\frac{7.1009+23.1009-23.7}{7.1009+23.1009-23.7+1}\) + \(\frac{1}{7.1009+23.1009-23.7+1}\) = \(\frac{7.1009+23.1009-23.7+1}{7.1009+23.1009-23.7+1}\) = 1.
\(\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{177}{236}+\frac{76}{236}-\frac{6}{236}\right)}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>\(\frac{\left(\frac{6734}{247}-\frac{6498}{247}\right).\frac{247}{236}}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>(3/4+x)*27/33=236/247*247/236=1
3/4+x=1:27/33=33/27
x=33/27-3/4=132/108-81/108
x=51/108
Vậy x=51/108
A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3-4}+\frac{1}{4-5}+\frac{1}{5-6}+\frac{1}{6-7}\right)\right]\)
A=\(\frac{7}{4}.\left[33.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
A=\(\frac{7}{4}.\frac{44}{7}\)
A=11
Like cho mình nha bài này viết mỏi tay lắm