\(^{\dfrac{2}{5}}\)+\(\dfrac{4}{10}\)= \(\dfrac{19}{10}\) là đúng hai sai
\(\dfrac{2}{5}\)+\(\dfrac{3}{2}\)=\(\dfrac{19}{10}\)là đúng hay sai câu vừa nãy mik ghi nhầm
đúng ghi Đ,sai ghi S
a)\(\dfrac{2}{3}\) của một nửa là \(\dfrac{1}{3}\)
b)\(\dfrac{1}{5}\) của \(\dfrac{1}{4}\) là \(\dfrac{1}{20}\)
c)Một nửa của \(\dfrac{1}{2}\) là \(\dfrac{1}{4}\)
d)\(\dfrac{2}{5}\) của \(\dfrac{4}{7}\) là \(\dfrac{7}{10}\)
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Bài 2: Đúng ghi Đ, sai ghi S.
a) \(\dfrac{13}{39};\dfrac{3}{5};\dfrac{22}{33}\) đều là phân số tối giản.
b) Phân số \(\dfrac{5}{10}\) có thể viết thành phép chia 10 : 5
bài 1:
7 ha=......m2
16ha=....m2
1km2=.....ha
\(\dfrac{1}{10}\)ha=.....m2
\(\dfrac{1}{4}\)ha=.....m2
\(\dfrac{1}{100}\)ha=.....m2
Bài 2:đúng ghi đ sai ghi s
54km2<540ha
71ha>80000m2
5m28dm2=5\(\dfrac{8}{10}\)m2
Bài 1
7ha=70000m2
16ha=160000m2
1km2=100ha
\(\dfrac{1}{10}\)ha=1000m2
\(\dfrac{1}{4}\)ha=2500m2
\(\dfrac{1}{100}\)ha=100m2
Bài 2
54km2<540ha S
71ha>80000m2 Đ
5m28dm2=\(5\dfrac{8}{10}\)m2 S
Goridano
Tổng của hai số đầu là
\(\dfrac{5}{12}\times2=\dfrac{10}{12}\left(1\right)\)
Tổng của 3 số đầu là:
\(\dfrac{19}{36}\times3=\dfrac{19}{12}\left(2\right)\)
Tổng của 4 số là:
\(\dfrac{143}{249}\times4=\dfrac{143}{60}\)
Từ (1) và (2), ta thấy số thứ 3 là: \(\dfrac{19}{12}-\dfrac{10}{12}=\dfrac{48}{60}=\dfrac{4}{5}\)
Từ (2) và (3), ta thấy số cuối là:
\(\left(\dfrac{3}{4}+\dfrac{4}{5}\right)\div2=\dfrac{31}{40}\)
Số đầu là:
\(\dfrac{31}{40}-\dfrac{1}{10}=\dfrac{20}{40}=\dfrac{1}{2}\)
Theo (1), số thứ 2 là:
\(\dfrac{10}{12}-\dfrac{1}{2}=\dfrac{4}{12}=\dfrac{1}{3}\)
Đáp số : \(\dfrac{1}{2};\dfrac{1}{3};\dfrac{3}{4};\dfrac{4}{5}\)
bài 2: 1, \(\left(\dfrac{5}{6}\right)^{10}.\left(\dfrac{3}{10}\right)^{10}\)2,\(\left(\dfrac{4}{7}\right)^{19}:\left(\dfrac{-12}{35}\right)^{19}\) 3,\(\left(\dfrac{-3}{7}\right)^7:\left(\dfrac{-3}{5}\right)\)
Lời giải:
1.
$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$
$=\frac{1}{4^{10}}$
2.
$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$
3.
$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$
1) \(\left(\dfrac{5}{6}\right)^{10}\cdot\left(\dfrac{3}{10}\right)^{10}\)
\(=\left(\dfrac{5}{6}\cdot\dfrac{3}{10}\right)^{10}\)
\(=\left(\dfrac{1}{4}\right)^{10}\)
2) \(\left(\dfrac{4}{9}\right)^{19}:\left(\dfrac{-12}{35}\right)^{19}\)
\(=\left(\dfrac{4}{9}:\dfrac{-12}{35}\right)^{19}\)
\(=\left(\dfrac{4}{9}\cdot\dfrac{35}{-12}\right)^{19}\)
\(=\left(-\dfrac{35}{27}\right)^{19}\)
3) \(\left(\dfrac{-3}{7}\right)^7:\left(\dfrac{-3}{5}\right)^7\)
\(=\left(\dfrac{-3}{7}:\dfrac{-3}{5}\right)^7\)
\(=\left(\dfrac{-3}{7}\cdot\dfrac{5}{-3}\right)^7\)
\(=\left(\dfrac{5}{7}\right)^7\)
1)\(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
2)\(\dfrac{ }{\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}}\)
3)\(\dfrac{ }{\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}}\)
4)\(\dfrac{ }{\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}}\)
1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
8. \(\dfrac{-5}{9}\) + \(\dfrac{8}{15}\) + \(\dfrac{-2}{11}\) + \(\dfrac{4}{-9}\) + \(\dfrac{7}{15}\)
9. \(\dfrac{2}{7}\) + (\(\dfrac{-2}{5}\) + \(\dfrac{5}{7}\))
10. \(\dfrac{7}{19}\). \(\dfrac{8}{11}\) + \(\dfrac{3}{11}\).\(\dfrac{7}{19}\)+\(\dfrac{-12}{19}\)
11. \(\dfrac{-5}{7}\).\(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\).\(\dfrac{9}{11}\)
12. \(\dfrac{-5}{13}\) + \(\dfrac{5}{7}\) + \(\dfrac{20}{41}\) + \(\dfrac{-8}{13}\) + \(\dfrac{21}{41}\)
Giúp tớ với ạ! Tớ cảm ơn! Các cậu chỉ cần ghi đáp án cuối cùng thôi ạ! Cảm ơn các cậu<3
8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)
9: =2/7-2/5+5/7=1-2/5=3/5
10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)
11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)
8 = -2/11
9 = 3/5
10 = -5/19
11 = -5/7
11 = 5/13
1 Đúng hoặc Sai,nếu sai thì sửa lại cho đúng
a/\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\) ; b/\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{10}\) ; c/\(\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-1\) ; d/\(\dfrac{8}{2\sqrt{8}-1}=\dfrac{P\left(2\sqrt{8}+1\right)}{4P-1}\) ; e/\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
2 Rút gọn các biểu thức
a/\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\) ; b/\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\) ; c/\(\dfrac{3+\sqrt{3}}{3-\sqrt{3}}+\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\) ; d/\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}+\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}}\)
Bài 2:
a) \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)
b) \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
c) \(\dfrac{3+\sqrt{3}}{3-\sqrt{3}}+\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\)
\(=\dfrac{\left(3+\sqrt{3}\right)^2+\left(3-\sqrt{3}\right)^2}{6}\)
\(=\dfrac{12+6\sqrt{3}+12-6\sqrt{3}}{6}=4\)
Bài 1:
a) Đúng
b) Sai vì \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
c) Sai vì \(\dfrac{2}{\sqrt{3}-1}=\sqrt{3}+1\)
e) Đúng