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a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)
\(=\frac{5-2^4.10}{2}\)
\(=5-8.10\)
\(=5-80\)
\(=-75\)
\(=\dfrac{-11}{5}\cdot\dfrac{-9}{11}\cdot\dfrac{-5}{4}\cdot\dfrac{2}{5}+2\cdot\dfrac{1}{4}\)
\(=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
b) \(\frac{x}{2}\)= \(\frac{y}{3}\) ; \(\frac{y}{5}\)= \(\frac{z}{7}\)và x+y+z=92
\(\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\)và x+y+z=92
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)và x+y+z=92
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}\)=\(\frac{92}{46}=2\)
Suy ra \(\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=2\Rightarrow z=42\)
Vậy ...
câu dưới tương tự nha bn
hoặc bn vào các câu hỏi tương tự ấy có nhiều bài dạng như vầy lắm
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
6:
\(=\dfrac{-8}{27}-3\cdot\dfrac{4}{9}+\dfrac{4}{3}+4\)
\(=-\dfrac{8}{27}+4=\dfrac{100}{27}\)
7: \(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right)\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left[\dfrac{117-236}{36}\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)
\(\dfrac{\left(-0.25\right)^{-5}\cdot9^4\left(-2\right)^{-3}-2^{-3}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)
\(=\dfrac{2^7\cdot3^8-2^6\cdot3^9}{2^9\cdot3^6+3^6\cdot2^9\cdot5}\)
\(=\dfrac{2^6\cdot3^8\left(2-3\right)}{2^9\cdot3^6\cdot6}\)
\(=\dfrac{1}{2^3}\cdot3^2\cdot\dfrac{-1}{6}\)
\(=\dfrac{-9}{6\cdot2^3}=\dfrac{-3}{2^4}=\dfrac{-3}{16}\)