Tính giá trị biểu thức:
A= 11 x 3^22 x 3^7 - 9^15
(2 x 30^14) ^ 2
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Sửa đề: A=7/12-9/20+11/30-13/42+15/56
A=1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8
=1/3+1/8=11/24
\(\frac{11\times3^{22}\times3^7\times9^{15}}{\left(2\times3^{14}\right)^2}=\frac{11\times3^{29}\times\left(3^2\right)^{15}}{2^2\times3^{28}}\)\(=\frac{11\times3^{29}\times3^{30}}{4\times3^{28}}=\frac{11\times3^{59}}{4\times3^{28}}\)\(=\frac{11\times3^{31}}{4}\)
- Kết quả to quá nên bạn tưn tính nhé !
a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145
b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7
c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11
a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21
b) (5/17 x 21/32 x 47/24 x 0)
= 0
c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7
a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x
b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15
c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2
a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36
câu,b,không,đủ,thông,tin,nhan,bạn.
a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{14}{5}\)
b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)
\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)
\(=\dfrac{9}{5}+1\)
\(=\dfrac{9}{5}+\dfrac{5}{5}\)
\(=\dfrac{14}{5}\)
\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{7}{5}-\dfrac{1}{6}\)
\(=\dfrac{42}{30}-\dfrac{5}{30}\)
\(=\dfrac{37}{30}\)
Bài 2:
Sau 2 giờ con ốc sên bò được:
1/5+1/6=11/30(cây cột)
Sau 2 giờ thì con ốc sên còn phải bò:
1-11/30=19/30(cây cột)
\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)
\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.30^{14}\right)^2}=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.30^{28}}=\frac{11.3^{29}-3^{30}}{2^2.\left(3.10\right)^{28}}=\frac{11.3^{29}-3.3^{29}}{2^2.10^{28}.3^{28}}\)
\(=\frac{3^{29}\left(11-3\right)}{2^2.10^{28}.3^{28}}=\frac{3.2}{10^{28}}=\frac{6}{10^{28}}\)