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a: \(=\dfrac{32}{9}+\dfrac{13}{6}=\dfrac{32\cdot2+13\cdot3}{18}=\dfrac{64+39}{18}=\dfrac{103}{18}\)
b: \(=\dfrac{43}{8}-\dfrac{43}{6}=\dfrac{-43}{24}\)
c:\(=4-2-\dfrac{1}{6}=2-\dfrac{1}{6}=\dfrac{11}{6}\)
d: \(=5+\dfrac{2}{3}+7+\dfrac{1}{2}-3-\dfrac{1}{2}+1+\dfrac{2}{3}\)
\(=10+\dfrac{4}{3}=\dfrac{34}{3}\)
a, \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(A=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(A=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(A=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
b, \(B=\dfrac{6}{15.18}+\dfrac{6}{18.21}+...+\dfrac{6}{87.90}\)
\(B=2\left(\dfrac{3}{15.18}+\dfrac{3}{18.21}+...+\dfrac{13}{87.90}\right)\)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(B=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)=2.\dfrac{1}{18}=\dfrac{1}{9}\)
c, \(C=\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
\(C=3\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(C=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)=3.\dfrac{3}{35}=\dfrac{9}{35}\)
Chúc bạn học tốt!!!
1. x3 - \(\dfrac{4}{25}\)x = 0
<=> x(x2 - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh
1 ) \(x^3-\dfrac{4}{25}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)
Vậy .............
2 ) \(3^{4x+4}=9^{x+2}\)
\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)
\(\Leftrightarrow4x+4=2x+4\)
\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)
3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )
4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)
\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)
\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)
Vậy ......
5) \(2+4+6+...+2x=210\)
\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)
\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)
\(\Leftrightarrow1+2+3+...+x=105\)
\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)
\(\Leftrightarrow x\left(x+1\right)=210\)
Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)
\(\Leftrightarrow x=14\)
Vậy ......
6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x=17.\)
Vậy ...........
\(\)
các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho
Có nhé bạn
Và ngoài ra \(-\frac{3}{6}=\frac{-3}{6}=\frac{3}{-6}\)
HT
TL :
Ko nhé
\(-\frac{3}{6}\)không phải số nào cả , nó không hề tồn tại ( \(x\notin N;x\in\varnothing\)
Vậy \(\frac{-3}{6}\)lớn hơn