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a: \(3200:40\cdot2=160\)
b: \(3920:28:2=70\)
c: \(\left(3^4\cdot5^7-9^2\cdot21\right):3^5=3^2\cdot5^7-7=703118\)
A = 7 (7 / 2.9 + 7 / 9.16 + .......... + 7/65.72)
A=7( 1/2 - 1/9 +1/9 - 1/16 +......+1/65 - 1/72)
A= 7 ( 1/2 -1/72)
A= 7 . 35/72
A=245/72
\(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7}{16.23}+.....+\frac{7^2}{65.72}\)
=\(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)
=\(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=\(7.\frac{35}{72}\)
=\(\frac{245}{72}\)
Đặt \(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+\frac{7^2}{23.30}\)
\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\right)\)
\(\Rightarrow A=7.\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(\Rightarrow A=\frac{49}{15}\)
đặt biểu thức là B
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có :
\(B=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+.....+\frac{1}{23}-\frac{1}{30}\right)\)
\(B=7.\left(\frac{1}{2}-\frac{1}{30}\right)=7.\frac{7}{15}=\frac{49}{15}\)
Ai thấy đúng thì ủng hộ nha !!!
\(a,x+\dfrac{5}{7}=\dfrac{10}{-35}\\ x=\dfrac{-10}{35}-\dfrac{5}{7}\\ x=-1\\ b,\dfrac{2}{x}=\dfrac{x}{8}\\ x.x=2.8\\ x^2=16\\ x=\pm4\)
a, \(x\) + \(\dfrac{5}{7}\) = \(\dfrac{10}{-35}\)
\(x\) = \(-\dfrac{10}{35}\) - \(\dfrac{5}{7}\)
\(x\) = -1
b, \(\dfrac{2}{x}\) = \(\dfrac{x}{8}\)
\(x^2\) = 16
\(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -4; 4}
c, \(\dfrac{x-3}{2}\) = \(\dfrac{32}{x-3}\)
(\(x-3\))(\(x-3\)) = 32\(\times\) 2
(\(x-3\))(\(x\) - 3) = 64
(\(x-3\))2 = 82
\(\left[{}\begin{matrix}x-3=-8\\x-3=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+3\\x=8+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=11\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -5; 11}
Ta có:
C = \(\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)
=> C = \(7.\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+...+\frac{7}{65.72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)
=> C = \(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)
=> C = \(7.\frac{35}{72}=\frac{245}{72}\)
Nhìn kĩ là ra thôi :
\(\frac{7^2}{2.9}+\frac{7^2}{9.16}+...+\frac{7^2}{65.72}\)
= \(7\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right)\)
= \(7\left(\frac{1}{2}-\frac{1}{72}\right)\)
= \(7.\frac{35}{72}=3\frac{29}{72}\)
Bài 2:
a) Ta có:
\(7\cdot2018\) ⇒ tích này chia hết cho 7
b) Ta có:
\(2020\cdot56\)
\(=2020\cdot8\cdot7\) ⇒ tích này chia hết cho 7
c) Ta có:
\(4\cdot23\cdot16\)
\(=2\cdot2\cdot23\cdot2\cdot2\cdot2\cdot2\) ⇒ tích này không chia hết cho 7
d) Ta có:
\(12\cdot8\cdot721\)
\(=12\cdot8\cdot103\cdot7\) ⇒ tích này chia hết cho 7
\(A=\frac{3^4.5^7-9^2.21}{3^5}=\frac{3^4.5^7-3^4.21}{3^5}=\frac{3^4.\left(5^7-21\right)}{3^5}=\frac{78125-21}{3}=\frac{78104}{3}\)
Vậy : \(A=\frac{78104}{3}\)
\(B=\frac{2^9.16+2^8.68}{2^{10}}=\frac{2^9.16+2^8.2.34}{2^{10}}=\frac{2^9.16+2^9.34}{2^{10}}\)
\(=\frac{2^9.\left(16+34\right)}{2^{10}}=\frac{2^9.50}{2^{10}}=\frac{50}{2}=25\)
Vậy :\(B=25\)