Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
1: \(=5^{20}\cdot\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)^8-1\)
=1+1-1=1
2: \(=\dfrac{15-8}{6}\cdot\dfrac{6}{7}+\left(-\dfrac{3}{2}\right)^2\)
=1+9/4
=13/4
3: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{3^8\cdot2^{10}+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{3^8\cdot2^{10}\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
a: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
b: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(-2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
a) \(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(5.2^2\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}=\frac{5^4.5^4.2^8}{5^{10}.2^{10}}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}.\)
Chúc bạn học tốt!
4^5=2^10
9^4=3^8
2*6^9=2^10*3^9
thì cái tử sẽ đc:
2^10*(-3)
mẫu e phân tích tt
Ta có
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\cdot\frac{5^4.20^4}{25^5.4^5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\cdot\frac{2^8.5^8}{5^{10}.2^{10}}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\cdot\frac{1}{5^2.2^2}\)
\(=\frac{\left(-2\right)}{6}\cdot\frac{1}{100}=-\frac{1}{3}\cdot\frac{1}{100}=-\frac{1}{300}\)
Vậy : \(E=-\frac{1}{300}\)
Bài làm
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)
\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{1}{300}\)
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
a) \(A=\frac{2^4.25^4}{10^5.5^5}=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}=\frac{2^4.5^8}{2^5.5^{10}}=\frac{1}{2.5^2}=\frac{1}{50}\)
b) \(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(B=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
\(A=\frac{2^4.25^4}{10^5.5^5}\)
\(A=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}\)
\(A=\frac{2^4.5^8}{2^5.5^5.5^5}\)
\(A=\frac{5^8}{2.5^{10}}\)
\(A=\frac{1}{2.5^2}=\frac{1}{2.25}=\frac{1}{50}\)
\(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(B=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(B=\frac{2^{10}.3^8-2^{10}.2.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(B=\frac{2^{10}.\left(3^8-2.3^9\right)}{2^{10}.\left(3^8+3^8.5\right)}\)
\(\Rightarrow B=\frac{3^8.2.3^9}{3^8+3^8.5}\) ( \(2^{10}\ne0\))
\(B=\frac{3^8.\left(1-2.3\right)}{3^8.\left(1+5\right)}\)
\(B=\frac{1-6}{1+5}\left(3^8\ne0\right)\)
\(B=\frac{-5}{6}\)
Tham khảo nhé~
Bài 5: GTNN chứ nhỉ?
Với mọi gt của \(x;y\in R\) ta có:
\(x^2+3\left|y-2\right|+1\ge1\)
Hay \(A\ge1\) với mọi gt của \(x;y\in R\)
Dấu "=" sảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Vậy..................
Bài 6: GTLN chứ?
Với mọi giá trị của \(x\in R\) ta có:
\(-\left(2x-1\right)^2\le0\Rightarrow-5-\left(2x-1\right)^2\le-5\)
Hay \(B\le5\) với mọi giá trị của \(x\in R\)
Dấu "=" sảy ra khi và chỉ khi \(x=\dfrac{1}{2}\)
Vậy...................
Bài 4 :
\(a,3^{15}-9^6=3^{15}-\left(3^2\right)^6=3^{15}-3^{12}=3^{12}\left(3^3-1\right)=3^{12}.26=3^{12}.2.13⋮\left(đpcm\right)\)
\(b,8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}=2^{18}\left(2^3-1\right)=2^{18}.7=2^{17}.2.7=2^{17}.14⋮14\left(đpcm\right)\)
Bài 5 :
\(A=1^2+3^2+6^2+9^2+.............+39^2\)
\(=1+3^2+\left(6^2+9^2+.........+39^2\right)\)
\(=10+3^2\left(2^2+3^2+.........+13^2\right)\)
\(=10+3^2.818\)
\(=10+9.818\)
\(=7372\)
2. Ta có
5x+5x+2=650 <=> 5x+5x.52=650 <=> 5x.(1+25)=650
<=> 5x.26=650
<=>5x=25=>x=2
3. TA CO: 55-54+53
=2375⋮7