\(|97\frac{2}{3}-125\frac{3}{5}|+97\frac{2}{5}-125\frac{1}{3}\)
Những câu hỏi liên quan
\(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
\(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{3}-125\frac{3}{5}\)
\(=\left|-\frac{419}{15}\right|+\left(-\frac{419}{15}\right)\)
\(=\frac{419}{15}+\left(-\frac{419}{15}\right)=0\)
học tốt ~~
Tính D= |97\(\frac{2}{3}\) -125 \(\frac{3}{5}\)|+97\(\frac{2}{5}\) -125\(\frac{1}{3}\)
Tính = cák hợp lý:
a) \(139\frac{5}{7}:\frac{2}{3}-138\frac{2}{7}:\sqrt{\frac{4}{9}}\)
b) \(\left(\frac{-5}{11}:\frac{13}{18}-\frac{5}{11}:\frac{13}{5}\right)+\frac{-1}{33}\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
a, \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\sqrt{\frac{4}{9}} \)
= \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\frac{2}{3}\)
= \((139\frac{5}{7}−138\frac{2}{7}):\frac{2}{3}\)
= \(1\frac{3}{7}:\frac{2}{3}\)
= \(2\frac{1}{7}\)
b, \((\frac{-5}{11}:\frac{13}{18}-\frac{5}{11}:\frac{13}{5})+\frac{-1}{33} \)
= \((\frac{5}{11}.\frac{-18}{13}-\frac{5}{11}.\frac{5}{13})+\frac{-1}{33}\)
= \([\frac{5}{11}.(\frac{-18}{13}-\frac{5}{13})]+\frac{-1}{33}\)
= \((\frac{5}{11}.\frac{-23}{13})+\frac{-1}{33}\)
= \(\frac{-155}{143}+\frac{-1}{33}\)
= \(\frac{-358}{429} \)
c, \(∣97\frac{2}{3}-125\frac{3}{5}∣+97\frac{2}{3}-125\frac{3}{5} \)
= \(∣\frac{-419}{15}∣+97\frac{2}{3}-125\frac{3}{5}\)
= \(\frac{419}{15}+97\frac{2}{3}-125\frac{3}{5}\)
= \(0\)
Tick cho mình nha!!!
Chúc bạn học tốt.
a) \(\frac{\left(-1\right)^3}{15}+\left(-\frac{2}{3}\right):2\frac{2}{3}-\left|-\frac{5}{6}\right|\)
b) \(1\frac{5}{13}-0,\left(3\right)-\left(1\frac{4}{9}+\frac{18}{13}-\frac{1}{3}\right)\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
d) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
tính\(\frac{\frac{125}{8}+\frac{125}{97}+\frac{125}{576}+\frac{250}{991}}{\frac{25}{8}+\frac{25}{97}+\frac{25}{576}+\frac{50}{991}}\)
\(\frac{\frac{125}{8}+\frac{125}{97}+\frac{125}{576}+\frac{250}{991}}{\frac{25}{8}+\frac{25}{97}+\frac{25}{576}+\frac{50}{991}}\)=\(\frac{250.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}{50.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}\)=\(\frac{250}{50}\)=5
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tính
\(P=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+...+\frac{1}{97}.3+\frac{1}{99}.1}\)
Lời giải:
** Sửa đề: Chỗ $\frac{1}{1}$ ở mẫu chuyển thành $\frac{1}{2}$
$\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+....+\frac{1}{97}.3+\frac{1}{99}.1$
$=50+(\frac{97}{3}+1)+(\frac{95}{5}+1)+....+(\frac{3}{97}+1)+(\frac{1}{99}+1)$
$=50+\frac{100}{3}+\frac{100}{5}+...+\frac{100}{97}+\frac{100}{99}$
$=100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})$
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})}=\frac{1}{100}\)
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\(|97\dfrac{2}{3}-125\dfrac{3}{5}|+97\dfrac{2}{5}-125\dfrac{1}{3}\)
Ta có: \(\left|97\dfrac{2}{3}-125\dfrac{3}{5}\right|+97\dfrac{2}{5}-125\dfrac{1}{3}\)
\(=\left|97+\dfrac{2}{3}-125-\dfrac{3}{5}\right|+97+\dfrac{2}{5}-125-\dfrac{1}{3}\)
\(=\left|-28+\dfrac{1}{15}\right|-28+\dfrac{1}{15}\)
\(=\left|\dfrac{1}{15}-28\right|-28+\dfrac{1}{15}\)
\(=28-\dfrac{1}{15}-28+\dfrac{1}{15}\)
\(=0\)
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giải PT: a, (4x-5)2 (2x-3)(x-1)=9
b,\(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}+\frac{2}{x+3}\)
c,(\(\left(\frac{x-1}{99}+\frac{x-99}{1}\right)+\left(\frac{x-3}{97}+\frac{x+97}{3}\right)+\left(\frac{x-5}{93}+\frac{x-95}{5}\right)=6\)
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
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Thuc hien phep tinh:
a/\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}\)+ \(\frac{0,6-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-0,16-\frac{4}{125}-\frac{4}{625}}\)
b/ \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
giúp tui với : -8 . 25 . ( -2 ) . ( -25 ) .4 .125
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