tính giá trị của biểu thức
a) \(log_575\)+ \(log_53\)
b) \(4log_{12}2\) + \(2log_{12}3\)
c) \(\dfrac{1}{3}log_3\dfrac{9}{7}+log_37^{\dfrac{1}{3}}\)
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\(log_2\dfrac{9}{10}+log_330=\) ? bạn chắc đề đúng chứ, 2 cơ số ko giống nhau, rút gọn cũng được nhưng nó sẽ không gọn trên thực tế.
\(log_3\dfrac{5}{9}-2log_3\sqrt{5}=log_3\dfrac{5}{9}-log_35=log_3\left(\dfrac{1}{9}\right)=log_33^{-2}=-2\)
\(log_2\dfrac{16}{3}+2log_2\sqrt{6}=log_2\dfrac{16}{3}+log_26=log_2\left(\dfrac{16}{3}.6\right)=log_232=log_22^5=5\)
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)
\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
TICK CHO MÌNH NHÉ
Giải:
\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)
= \(\dfrac{8}{9}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
`a)100x^2-20x+1`
`=(10x-1)^2`
Thay `x=1/10`
`=>100x^2-20x+1=(1-1)^2=0`
`b)49x^2-42x+10`
`=49*4/49-42*2/7+10`
`=4-12+10=2`
`c)25x^2+40x+16y^2`
`=(5x+4y)^2=(2+3)^2=25`
a: \(A=2\cdot2^2-\dfrac{1}{3}\cdot9=8-3=5\)
b: \(B=\dfrac{1}{2}a^2-3b^2=\dfrac{1}{2}\cdot4-3\cdot\dfrac{1}{9}=2-\dfrac{1}{3}=\dfrac{5}{3}\)
\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)
\(log_575+log_53=log_5\left(75.3\right)=log_5225\)
\(4log_{12}2+2log_{12}3=log_{12}16+log_{12}9=log_{12}\left(16.9\right)=log_{12}144=log_{12}12^2=2\)
\(\dfrac{1}{3}log_3\dfrac{9}{7}+log_37^{\dfrac{1}{3}}=\dfrac{1}{3}\left(log_3\dfrac{9}{7}+log_37\right)=\dfrac{1}{3}log_3\left(\dfrac{9}{7}.7\right)=\dfrac{1}{3}log_39=\dfrac{2}{3}\)