1) \(\dfrac{12}{24}=\dfrac{12}{2\times12}=\dfrac{1}{2}\) ; \(\dfrac{18}{48}=\dfrac{3\times6}{6\times8}=\dfrac{3}{8}\) ;\(\dfrac{72}{48}=\dfrac{24\times3}{24\times2}=\dfrac{3}{2}\) ; \(\dfrac{1212}{2424}=\dfrac{1212}{1212\times2}=\dfrac{1}{2}\) ; \(\dfrac{132639}{142842}=\dfrac{13\times10203}{14\times10203}=\dfrac{13}{14}\) ;

2)

a) \(\dfrac{7}{9}=\dfrac{7\times8}{9\times8}=\dfrac{56}{72}và\dfrac{5}{8}=\dfrac{5\times9}{8\times9}=\dfrac{45}{72}\)

b) \(\dfrac{8}{21}và\dfrac{4}{7}=\dfrac{4\times3}{7\times3}=\dfrac{12}{21}\)

c) \(\dfrac{2}{3}=\dfrac{2\times5}{3\times5}=\dfrac{10}{15};\dfrac{4}{5}=\dfrac{4\times3}{5\times3}=\dfrac{12}{15}và\dfrac{13}{15}\) 

3) \(\dfrac{6}{7}và\dfrac{5}{7}\) \(6>5\) \(nên\) \(\dfrac{6}{7}>\dfrac{5}{7}\);

 \(\dfrac{9}{24}và\dfrac{5}{8}\) \(có:\) \(\dfrac{9}{24}=\dfrac{3}{8}\) \(3< 5\) \(nên\) \(\dfrac{3}{8}< \dfrac{5}{8}hay\dfrac{9}{24}< \dfrac{5}{8}\) ;

\(\dfrac{5}{7}và\dfrac{7}{9}\) \(có:\) \(\dfrac{5}{7}=\dfrac{5\times9}{7\times9}=\dfrac{45}{63}\) \(và\) \(\dfrac{7}{9}=\dfrac{7\times7}{9\times7}=\dfrac{49}{63}\) \(45< 49\) \(nên\) \(\dfrac{45}{63}< \dfrac{49}{63}hay\dfrac{5}{7}< \dfrac{7}{9}\);

\(\dfrac{11}{24}và\dfrac{11}{32}\) \(32>24\) \(nên\) \(\dfrac{11}{24}>\dfrac{11}{32}\)

\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

Ok

\(\left\{{}\begin{matrix}6u_2+u_5=1\\3u_3+2u_4=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6u_1.q+u_1.q^4=1\\3u_1.q^2+2u_1.q^3=-1\end{matrix}\right.\)

\(\Rightarrow u_1\left(6q+q^4+3q^2+2q^3\right)=0\)

\(\Leftrightarrow q^3+2q^2+3q+6=0\)

\(\Leftrightarrow\left(q+2\right)\left(q^2+3\right)=0\)

\(\Leftrightarrow q=-\text{​​}2\)

\(\Rightarrow u_1=\dfrac{1}{4}\)

\(\Rightarrow u_n=u_1.q^{n-1}=\dfrac{1}{4}.\left(-2\right)^{n-1}=\left(-2\right)^{n-3}\)

\(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne9\right)\)

Để \(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}< 0\) thì 

\(\sqrt{x}-3< 0\) ( do \(\sqrt{x}+3\ge3>0\))

\(\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

Để M là số nguyên âm thì \(\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay \(x\in\left\{0;1;4\right\}\)

\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

a.

Hệ có nghiệm duy nhất khi:

\(\dfrac{m}{2}\ne\dfrac{1}{-1}\Rightarrow m\ne-2\)

b.

Hệ có vô số nghiệm khi:

\(\dfrac{1}{1}=\dfrac{m}{-1}=\dfrac{3}{3}\Rightarrow m=-1\)

c.

Hệ vô nghiệm khi:

\(\dfrac{2}{-4}=\dfrac{-1}{2}\ne\dfrac{-m}{4}\Rightarrow m\ne2\)