Tính và so sánh :
a) \(\left(\frac{-2}{3}\right)^3\) và \(\frac{\left(-2\right)^3}{3^3}\)
b) \(\frac{10^5}{2^5}\)và \(\left(\frac{10}{2}\right)^5\)
tính và so sánh. giúp mình nha
\(\left(\frac{_{-2}}{3}\right)^3\)và \(\frac{\left(-2\right)^3}{3^3}\)
\(\frac{10^5}{2^5}\)và \(\left(\frac{10}{2}\right)^5\)
\(\left(\frac{-2}{3}\right)^3=\frac{-8}{27}\)
\(\frac{\left(-2\right)^3}{3^3}=\frac{-8}{27}\)
\(=>\left(-\frac{2}{3}\right)^3=\frac{\left(-2\right)^3}{3^3}\)
Áp dụng câu trên ta được :
\(\frac{10^5}{2^5}=\left(\frac{10}{2}\right)^5\)
Ủng hộ nha
o0o I am a studious person o0o:cái này có công thức trong sách giáo khoa rồi cần j phải tính
so sánh :
C=\(\frac{5^4.20^4}{25^5.4^5}\) và D=\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
E=\(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)và F=\(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3\)
I=\(\frac{2}{3}+\frac{1}{3}:\left(\frac{-8}{25}\right)\)và H=\(\frac{5}{11}.\frac{4}{11}+\frac{7}{11}.\frac{5}{11}-\frac{2}{3}\)
Tính và so sánh :
a) (2 .5)2 và 22 . 52
b) \(\left(\frac{1}{2}.\frac{3}{4}\right)^3\) và \(\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
Theo bài ra , ta có :
\(\left(2.5\right)^2=10^2\)
\(2^2.5^2=\left(2.5\right)^2=10^2\)
Vì \(10^2=10^2=100\)
Vậy \(\left(2.5\right)^2=2^2.5^2\)
b)
\(\left(\frac{1}{2}.\frac{3}{4}\right)^3=\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
mà \(\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\) là vế phải
Vậy \(\left(\frac{1}{2}.\frac{3}{4}\right)^3=\left(\frac{1}{2}\right)^3.\left(\frac{3}{4}\right)^3\)
bài 1: tính
a) \(\left(\frac{3}{7}\right)^0=\frac{7}{9}:\left(\frac{2}{3}\right)^2-\left|-\frac{4}{5}\right|\)
b) \(\frac{10^3+2.5^3+5^3}{55}\)
bài 2: so sánh
\(^{3^{2009}}\) và \(9^{1005}\)
thanksss~~~
a.
\(\left(\frac{3}{7}\right)^0+\frac{7}{9}\div\left(\frac{2}{3}\right)^2-\left|-\frac{4}{5}\right|=0+\frac{7}{9}\div\frac{4}{9}-\frac{4}{5}=\frac{7}{9}\times\frac{9}{4}-\frac{4}{5}=\frac{7}{4}-\frac{4}{5}=\frac{35}{20}-\frac{16}{20}=\frac{19}{20}\)
b.
\(\frac{10^3+2\times5^3+5^3}{55}=\frac{\left(2\times5\right)^3+2\times5^3+5^3}{55}=\frac{2^3\times5^3+2\times5^3+5^3}{5\times11}=\frac{5^3\times\left(2^3+2+1\right)}{5\times11}=\frac{5^2\times11}{11}=5^2=25\)
c.
\(3^{2009}< 3^{2010}=\left(3^2\right)^{1005}=9^{1005}\)
Vậy 32009 < 91005
Chúc bạn học tốt ^^
Cho A = \(\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)và B = \(\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
So sánh A và B
Ta có
\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\) \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\) \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{4968}{2225}\) \(\Leftrightarrow B=\frac{25}{8}\)
\(\Leftrightarrow A=\frac{39744}{17800}\) \(\Leftrightarrow B=\frac{55625}{17800}\)
Ta có: 39744<55625
\(\Rightarrow A< B\)
Vậy A<B
Thực hiện phép tính:
a,\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
b,\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
c,\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
Tính:
a)\(\frac{3}{5}+\left(\frac{-4}{3}\right)+\left(\frac{-3}{4}\right)\)
b) \(\frac{5}{8}-\left(\frac{-2}{5}\right)-\frac{3}{10}\)
c) \(\left(\frac{-7}{2}\right)+\left(\frac{-2}{3}\right)+\left(\frac{-3}{5}\right)\)
d) \(\frac{3}{4}-[\left(\frac{-5}{3}\right)-\left(\frac{1}{12}+\frac{2}{9}\right)]\)
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
Cho x = 3,7.So sánh :
\(A=\left[x\right]+\left[x+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)
và B = [5x]
\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)
\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)
\(=5x+2\)
\(B=5x\)
\(\Rightarrow A>B\)Với \(\forall\)\(x\)
#)Giải :
\(A=\left[x\right]+\left[1+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)
Thay x = 3,7 vào biểu thức, ta có :
\(A=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)
\(A=\left[3,7+3,7+3,7+3,7+3,7\right]+\left[1+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right]\)
\(A=18,5+3\)
\(A=21,5\)
\(B=\left[5x\right]=\left[5\times3,7\right]=18,5\)
Vì 21,5 > 18,5 \(\Rightarrow A>B\)
Phạm Thị Thùy Linh+๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ ( Team TST 14 ):Cả 2 bạn đều nhầm chỗ \(\left[a\right]\) rồi nha.\(\left[a\right]\) tức là phần nguyên của a nghĩa là số nguyên lớn nhất ko vượt quá a.
\(A=\left[x\right]+\left[x+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)
\(=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)
\(=3+3+4+4+4\)
\(=18\)
\(B=\left[5x\right]\)
\(B=\left[18,5\right]\)
\(=18\)
Vậy \(A=B\left(=18\right)\)