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b: \(\dfrac{2}{5}-\left(\dfrac{4}{3}+\dfrac{4}{5}\right)-\left(-\dfrac{1}{9}-0,4\right)+\dfrac{11}{9}\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{4}{5}+\dfrac{1}{9}+\dfrac{2}{5}+\dfrac{11}{9}\)
\(=\left(\dfrac{2}{5}-\dfrac{4}{5}+\dfrac{2}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=-\dfrac{4}{3}+\dfrac{12}{9}=0\)
c: \(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{5}{13}+\dfrac{-2}{11}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{2}{11}\right]+\dfrac{3}{4}\)
\(=\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}=-\dfrac{7}{8}+\dfrac{3}{4}=-\dfrac{1}{8}\)
a:\(\widehat{BAC}+\widehat{xAC}=180^0\)(hai góc kề bù)
=> \(\widehat{BAC}+70^0=180^0\)
=>\(\widehat{BAC}=110^0\)
Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)
mà hai góc này là hai góc ở vị trí trong cùng phía
nên AC//BD
b: Vì AC//BD
nên \(\widehat{yCx}=\widehat{CDB}\)(hai góc đồng vị)
=>\(\widehat{yCx}=60^0\)
Ta có: \(\widehat{yCx}+\widehat{ACD}=180^0\)(hai góc kề bù)
=>\(\widehat{ACD}+60^0=180^0\)
=>\(\widehat{ACD}=120^0\)
Ta có: \(\widehat{BAC}+\widehat{ABD}=180^0\)(AC//BD)
=>\(\widehat{BAC}+70^0=180^0\)
=>\(\widehat{BAC}=110^0\)
e: \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)
=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
=>x+1=0
=>x=-1
\(\left\{{}\begin{matrix}3x-4y=-2\\5x+2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-4y=-2\\2y=14-5x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-2\cdot2y=-2\\2y=14-5x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(14-5x\right)=-2\\2y=14-5x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x-28+10x=-2\\2y=-5x+14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=-2+28=26\\2y=-5x+14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\2y=-5\cdot2+14=14-10=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Bài 8:
\(4)-\dfrac{20}{7}:\dfrac{5}{21}\\ =-\dfrac{20}{7}\cdot\dfrac{21}{5}\\ =-4\cdot3\\ =-12\\ 5)-\dfrac{8}{5}:\dfrac{-12}{7}\\ =\dfrac{-8}{5}\cdot\dfrac{-7}{12}\\ =\dfrac{14}{15}\\ 6)\dfrac{-12}{21}:\dfrac{1}{6}\\ =\dfrac{-4}{7}\cdot6\\ =-\dfrac{24}{7}\)
Bài 10:
1: \(4,5\cdot\left(-\dfrac{4}{9}\right)=-\dfrac{9}{2}\cdot\dfrac{4}{9}=-\dfrac{4}{2}=-2\)
2: \(2,4\cdot\left(-3\dfrac{4}{7}\right)=\dfrac{-12}{5}\cdot\dfrac{25}{7}=\dfrac{-12}{7}\cdot\dfrac{25}{5}=-5\cdot\dfrac{12}{7}=-\dfrac{60}{7}\)
3: \(0,2\cdot\dfrac{-15}{4}=\dfrac{1}{5}\cdot\dfrac{-15}{4}=-\dfrac{15}{5}\cdot\dfrac{1}{4}=-\dfrac{3}{4}\)
4: \(\left(-3,5\right):\left(-2\dfrac{4}{5}\right)=\dfrac{3.5}{2\dfrac{4}{5}}=\dfrac{3.5}{2.8}=\dfrac{5}{4}\)
5: \(\dfrac{-5}{23}:\left(-2\right)=\dfrac{5}{23}:2=\dfrac{5}{23\cdot2}=\dfrac{5}{46}\)
6: \(1,25:\left(-3\dfrac{1}{8}\right)=1.25:\dfrac{-25}{8}=1.25\cdot\dfrac{-8}{25}=-\dfrac{10}{25}=-\dfrac{2}{5}\)
Bài 9:
1: \(-3\dfrac{1}{9}\cdot\dfrac{4}{21}=\dfrac{-28}{9}\cdot\dfrac{4}{21}=-\dfrac{28}{21}\cdot\dfrac{4}{9}=-\dfrac{4}{9}\cdot\dfrac{4}{3}=-\dfrac{16}{27}\)
2: \(-\dfrac{3}{4}\cdot2\dfrac{1}{2}=-\dfrac{3}{4}\cdot\dfrac{5}{2}=\dfrac{-15}{8}\)
3: \(-\dfrac{8}{15}\cdot1\dfrac{1}{4}=-\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{-40}{60}=-\dfrac{2}{3}\)
4: \(-\dfrac{11}{15}:1\dfrac{1}{10}=-\dfrac{11}{15}:\dfrac{11}{10}=-\dfrac{11}{15}\cdot\dfrac{10}{11}=-\dfrac{10}{15}=-\dfrac{2}{3}\)
5: \(1\dfrac{1}{5}:\left(-2\dfrac{1}{5}\right)=\dfrac{6}{5}:\dfrac{-11}{5}=\dfrac{6}{5}\cdot\dfrac{5}{-11}=\dfrac{6}{-11}=-\dfrac{6}{11}\)
6: \(\left(-3\dfrac{1}{7}\right):\left(-1\dfrac{6}{49}\right)=\dfrac{-22}{7}:\dfrac{-55}{49}=\dfrac{22}{7}\cdot\dfrac{49}{55}\)
\(=\dfrac{22}{55}\cdot\dfrac{49}{7}=7\cdot\dfrac{2}{5}=\dfrac{14}{5}\)
\(6)\left(2x+1\right)^2-\left(x+3\right)^2=0\\ \Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\\ 7)\left(x^2-4\right)+x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)+x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ 8)2\left(x+1\right)=\left(5x-1\right)\left(x+1\right)\\ \Leftrightarrow2\left(x+1\right)-\left(5x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2-5x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(3-5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{5}\end{matrix}\right.\\ 9)\left(-4x+3\right)x=\left(2x+5\right)x\\ \Leftrightarrow\left(-4x+3\right)x-\left(2x+5\right)x=0\\ \Leftrightarrow x\left(-4x+3-2x-5\right)=0\\ \Leftrightarrow x\left(-6x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: \(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot5^n\)
=>\(2^n\cdot\left(\dfrac{1}{2}+4\right)=5^n\cdot9\)
=>\(2^n\cdot\dfrac{9}{2}=5^n\cdot9\)
=>\(2^{n-1}=5^n\)
=>\(n-1=n\cdot log_25\)
=>\(n\left(1-log_25\right)=1\)
=>\(n=\dfrac{1}{1-log_25}\)
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\\ \Rightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+3\right)=0\\ \Rightarrow\dfrac{x-100}{30}+\dfrac{x-100}{43}+\dfrac{x-100}{95}+\dfrac{x-100}{8}=0\\ \Rightarrow\left(x-100\right)\left(\dfrac{1}{30}+\dfrac{1}{43}+\dfrac{1}{95}+\dfrac{1}{8}\right)=0\\ \Rightarrow x-100=0\\ \Rightarrow x=100\)
a; (2\(x\) - 1)2 = 49
(2\(x\) - 1)2 = 72
\(\left[{}\begin{matrix}2x-1=-7\\2x-1=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-7+1\\2x=7+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-6\\2x=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{6}{2}\\x=\dfrac{8}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- 3; 4}
b; (5\(x-3\))2 - (4\(x\) - 7)2 = 0
(5\(x\) - 3 - 4\(x\) + 7)(5\(x\) - 3 + 4\(x\) - 7) = 0
(\(x\) + 4)(9\(x\) - 10) = 0
\(\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- 4; \(\dfrac{10}{9}\)}