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Bài 1:
a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)
b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=1\)
Bài 2:
a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)
b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)
c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)
d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)
Bài 3:
a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)
\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)
\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)
\(=9.\dfrac{2}{3}.-1\)
\(=6.-1=-6\)
b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
a,
\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)
\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)
b,
\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
c,
\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)
\(\Rightarrow5\left|x+1\right|^2=180\)
\(\Rightarrow\left|x+1\right|^2=36\)
Mà \(\left|x+1\right|\ge0\)
=> x + 1 = 6 <=> x = 7
\(\left|\dfrac{7}{8}x+\dfrac{5}{6}\right|-\left|\dfrac{1}{2}x+5\right|=0\\ \Leftrightarrow\left|\dfrac{7}{8}x+\dfrac{5}{6}\right|=\left|\dfrac{1}{2}x+5\right|\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x+\dfrac{5}{6}=\dfrac{1}{2}x+5\\\dfrac{7}{8}x+\dfrac{5}{6}=-\left(\dfrac{1}{2}x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{100}{9}\\-\dfrac{140}{33}\end{matrix}\right.\)
\(\text{a) }\left(x-1\right)\left(x-5\right)>0\\ \text{ Để }\left(x-1\right)\left(x-5\right)>0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ cùng dấu }\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên dương }\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5>0\Rightarrow x>5\end{matrix}\right.\Rightarrow x>5\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên âm }\Rightarrow\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow x< 1\\ \text{Vậy }\left(x-1\right)\left(x-5\right)>0\text{ khi }x< 1\text{ hoặc }x>5\)
\(\text{b) }\left(x-1\right)\left(x-5\right)< 0\\ \text{ Để }\left(x-1\right)\left(x-5\right)< 0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ trái dấu }\\ \text{ Mà }x-1>x-5\\ \Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow1< x< 5\\ \text{ Vậy }\left(x-1\right)\left(x-5\right)< 0\text{ khi }1< x< 5\)
\(\text{c) }\dfrac{3}{4}-\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left|x-\dfrac{1}{7}\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{7}=-2\\x-\dfrac{1}{7}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{7}\\x=\dfrac{15}{7}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{13}{7}\text{ hoặc }x=\dfrac{15}{7}\)
\(\text{d) }\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=-\dfrac{1}{4}\\x-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{1}{4}\text{ hoặc }x=\dfrac{3}{4}\)
\(\text{e) }8\left(x+1\right)-2\left(2x+5\right)=0\\ \Leftrightarrow8x+8-4x+10=0\\ \Leftrightarrow\left(8x-4x\right)+\left(8+10\right)=0\\ \Leftrightarrow4x+18=0\\ \Leftrightarrow4x=-18\\ \Leftrightarrow x=-\dfrac{9}{2}\\ \text{Vậy }x=-\dfrac{9}{2}\)
\(\text{g) }\left(6x-1\right)-\left(x+8\right)=0\\ \Leftrightarrow6x-1-x-8=0\\ \Leftrightarrow\left(6x-x\right)-\left(1+8\right)=0\\ \Leftrightarrow5x-9=0\\ \Leftrightarrow5x=9\\ \Leftrightarrow x=\dfrac{9}{5}\\ \text{Vậy }x=\dfrac{9}{5}\)
\(\text{h) }\left|7x-\dfrac{1}{4}\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}7x-\dfrac{1}{4}=-1\\7x-\dfrac{1}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-\dfrac{3}{4}\\7x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{28}\\x=\dfrac{5}{28}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{3}{28}\text{ hoặc }x=\dfrac{5}{28}\)
\(\text{q) }-2x-3=-x+7\\ \Leftrightarrow-2x-3-\left(-x+7\right)=0\\ \Leftrightarrow-2x-3+x-7=0\\ \Leftrightarrow\left(-2x+x\right)-\left(3+7\right)=0\\ \Leftrightarrow-x-10=0\\ \Leftrightarrow-x=10\\ \Leftrightarrow x=-10\\ \text{ Vậy }x=-10\)
Ta có :
\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right)\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right)\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\\\dfrac{1}{3}x+\dfrac{4}{3}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=-4\end{matrix}\right.\)
Vậy : .......
\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right)\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right)\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\\\dfrac{1}{3}x+\dfrac{4}{3}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{7}x=\dfrac{2}{7}\\-\dfrac{1}{5}x=-\dfrac{3}{5}\\\dfrac{1}{3}x=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=-4\end{matrix}\right.\)
\(\dfrac{3}{8}+\left(x-1\dfrac{1}{7}\right)=0\)
\(\Rightarrow x-\dfrac{8}{7}=-\dfrac{3}{8}\)
\(\Rightarrow x=\left(-\dfrac{3}{8}\right)+\dfrac{8}{7}\)
\(\Rightarrow x=\dfrac{43}{56}\)