7. Tính giá trị của các biểu thức
A = -1,6 : (1+\(\frac{2}{3}\)); B= 1,4 . \(\frac{15}{49}\) - (\(\frac{4}{5}\)+\(\frac{2}{3}\)) : \(2\frac{1}{5}\).
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a. 1/3 + 1/4 - 1/6
= 7/12 - 1/6
= 5/12
b. 2/5 x 5/7 : 3/4
= 2/7 : 3/4
= 8/21
\(a,\) Số số hạng là \(\left(40-2\right):2+1=20\left(số\right)\)
Tổng là \(\left(40+2\right)\times20:2=420\)
\(b,\) Số số hạng là \(\left(39-1\right):2+1=20\left(số\right)\)
Tổng là \(\left(39+1\right)\times20:2=400\)
a) 2 + 4 + 6 + 8 + ... + 34 + 36 + 38 + 40
= ( 2 + 42 ) + ( 4 + 38 ) + .... + ( 20 + 22 )
= 42 \(\times\) 10
= 420
b) 1 + 3 + 5 + 7 + ... + 35 + 37 + 39
= ( 1 + 39 ) + ( 3 + 37 ) + ...+ ( 19 + 21 )
= 40 \(\times\) 10
= 400
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
a) \(\dfrac{1}{3}+\dfrac{4}{3}\times\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{4}{6}=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
b) \(\dfrac{3}{5}\times\dfrac{4}{7}:\dfrac{16}{21}=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{21}{16}=\dfrac{12}{35}\times\dfrac{21}{16}=\dfrac{252}{560}=\dfrac{9}{20}\)
\(log_575+log_53=log_5\left(75.3\right)=log_5225\)
\(4log_{12}2+2log_{12}3=log_{12}16+log_{12}9=log_{12}\left(16.9\right)=log_{12}144=log_{12}12^2=2\)
\(\dfrac{1}{3}log_3\dfrac{9}{7}+log_37^{\dfrac{1}{3}}=\dfrac{1}{3}\left(log_3\dfrac{9}{7}+log_37\right)=\dfrac{1}{3}log_3\left(\dfrac{9}{7}.7\right)=\dfrac{1}{3}log_39=\dfrac{2}{3}\)
a)
\(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)
b)
\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)
Theo vi et: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2020}{1}=-2020\\x_1x_2=\dfrac{c}{a}=\dfrac{2021}{1}=2021\end{matrix}\right.\)
a
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-2020}{2021}\)
b
\(x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2=\left(x_1+x_2\right)^2-2x_1x_2=\left(-2020\right)^2-2.2021=4076358\)
a) 27 + 34 + 66 = 27 + (34 + 66) = 27 + 100 = 127 | b) 7 x 5 x 2 = 7 x (5 x 2) = 7 x 10 = 70 |
`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`
giúp mình với huhu
lần sau giúp okm