a: \(\left(\dfrac{2}{3}\right)^5=\dfrac{2^5}{3^5}=\dfrac{32}{243}\)

\(\left(-\dfrac{2}{3}\right)^5=\dfrac{\left(-2\right)^5}{3^5}=\dfrac{-32}{243}\)

\(\left(-1\dfrac{3}{4}\right)^2=\left(-\dfrac{7}{4}\right)^2=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\)

\(\left(-0,1\right)^4=\left(0,1\right)^4=\left(\dfrac{1}{10}\right)^4=\dfrac{1}{10^4}=\dfrac{1}{10000}\)

b: \(\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)

\(\dfrac{790^4}{79^4}=\left(\dfrac{790}{79}\right)^4=10^4=10000\)

\(\dfrac{3^2}{15^2}=\left(\dfrac{3}{15}\right)^2=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)

\(\dfrac{\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^{n-1}}=\left(-\dfrac{1}{2}\right)^{n-n+1}=\left(-\dfrac{1}{2}\right)^1=-\dfrac{1}{2}\)

\(\dfrac{3}{4}x-6< =0\)

=>\(\dfrac{3}{4}x< =6\)

=>\(x< =6:\dfrac{3}{4}=6\cdot\dfrac{4}{3}=8\)

không hiểu chỗ nào cứ hỏi nha

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

mọi người giúp e với ạ

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

Khi x=5 thì \(11x-52=11\cdot5-52=55-52=3>0\)

=>Đúng

Khi x=5 thì \(6x-29=6\cdot5-29=30-29=1>0\)

=>6x-29>0 đúng

Khi x=5 thì 5-2=3<=0(sai)

=>x-2<=0 là đáp án sai duy nhất, hai cái còn lại đúng

Câu 1:

a: \(-\dfrac{9}{51}\cdot\dfrac{17}{6}=\dfrac{-9}{6}\cdot\dfrac{17}{51}=\dfrac{-3}{2}\cdot\dfrac{1}{3}=-\dfrac{1}{2}\)

b: \(-\dfrac{25}{32}\cdot\left(-0,2\right)=\dfrac{25}{32}\cdot0,2=\dfrac{5}{32}\)

c: \(-15,2\cdot3,5=-53,2\)

d: \(-\dfrac{8}{15}\cdot1\dfrac{1}{4}=-\dfrac{8}{15}\cdot\dfrac{5}{4}=-\dfrac{8}{4}\cdot\dfrac{5}{15}=-\dfrac{2}{3}\)

e: \(1\dfrac{2}{5}\cdot\dfrac{-3}{14}=\dfrac{7}{5}\cdot\dfrac{-3}{14}=-\dfrac{21}{70}=-\dfrac{3}{10}\)

f: \(1\dfrac{1}{17}\cdot1\dfrac{1}{36}=\dfrac{18}{17}\cdot\dfrac{37}{36}=\dfrac{18}{36}\cdot\dfrac{37}{17}=\dfrac{37}{17}\cdot\dfrac{1}{2}=\dfrac{37}{34}\)

Câu 2:

a: \(-\dfrac{5}{2}:\dfrac{5}{8}=-\dfrac{5}{2}\cdot\dfrac{8}{5}=-\dfrac{8}{2}=-4\)

b: \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)=\dfrac{-21}{5}:\dfrac{14}{5}=-\dfrac{21}{5}\cdot\dfrac{5}{14}=-\dfrac{21}{14}=-\dfrac{3}{2}\)

c: \(7:\left(-3,5\right)=-\dfrac{7}{3,5}=-2\)

d: \(-1\dfrac{4}{5}:\dfrac{-3}{4}=\dfrac{-9}{5}\cdot\dfrac{-4}{3}=\dfrac{-9}{3}\cdot\dfrac{-4}{5}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}\)

e: \(4,2:\dfrac{-15}{12}=4,2:\dfrac{-5}{4}=4,2\cdot\dfrac{-4}{5}=-\dfrac{16.8}{5}=-3,36\)

f: \(6\dfrac{9}{11}:\left(-3\right)=-\dfrac{75}{11}\cdot\dfrac{1}{3}=-\dfrac{25}{11}\)

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

3) Với a = 3 ta có hpt:

\(\left\{{}\begin{matrix}2x+3y=5\\3x+2y=2\cdot3+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+9y=15\\6x+4y=14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5y=1\\2x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\2x+\dfrac{3}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\2x=5-\dfrac{3}{5}=\dfrac{22}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=\dfrac{11}{5}\end{matrix}\right.\)

2: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\y\ne1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}-\dfrac{2}{x-2}+\dfrac{3}{y-1}=2-1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}=1-\dfrac{1}{y-1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y-1=5\\x-2=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=2+\dfrac{5}{4}=\dfrac{8}{4}+\dfrac{5}{4}=\dfrac{13}{4}\end{matrix}\right.\left(nhận\right)\)

Bài 6:

\(a)P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{33\cdot37}+\dfrac{2}{37\cdot41}\\ =\dfrac{1}{2}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{33\cdot37}+\dfrac{4}{37\cdot41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{33}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{41}\right)\\ =\dfrac{1}{2}\cdot\dfrac{40}{41}\\ =\dfrac{20}{41}\\ b)Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\\ =\dfrac{6}{7}\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{121}\right)\\ =\dfrac{6}{7}\cdot\dfrac{119}{242}\\ =\dfrac{51}{121}\)

Bài 5:

a: Để A>0 thì \(\dfrac{2a-1}{-5}>0\)

=>2a-1<0

=>\(a< \dfrac{1}{2}\)

b: Để A<0 thì \(\dfrac{2a-1}{-5}< 0\)

=>2a-1>0

=>2a>1

=>\(a>\dfrac{1}{2}\)

c: Để A=0 thì \(\dfrac{2a-1}{-5}=0\)

=>2a-1=0

=>2a=1

=>\(a=\dfrac{1}{2}\)

Bài 6:

a: \(P=\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+...+\dfrac{2}{37\cdot41}\)

\(=\dfrac{2}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{37\cdot41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{41}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{41}\right)=\dfrac{1}{2}\cdot\dfrac{40}{41}=\dfrac{20}{41}\)

b: \(Q=\dfrac{6}{2\cdot9}+\dfrac{6}{9\cdot16}+...+\dfrac{6}{114\cdot121}\)

\(=\dfrac{6}{7}\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{7}{114\cdot121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{114}-\dfrac{1}{121}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{1}{2}-\dfrac{1}{121}\right)=\dfrac{6}{7}\cdot\dfrac{119}{242}=\dfrac{51}{121}\)