a: Ta có: \(x^2+2xy+y^2-4x^2y^2\)

\(=\left(x+y\right)^2-4x^2y^2\)

\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)

b: Ta có: \(x^6-y^6\)

\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\)

c: Ta có: \(25-a^2+2ab-b^2\)

\(=25-\left(a-b\right)^2\)

\(=\left(5-a+b\right)\left(5+a-b\right)\)

\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)

\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)

Câu b xem lại đề đi nhé

cảm ơn bạn nhé

\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)

\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)

\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)

\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)

\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)

\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\)         ( 4^3 = 8^2 ; 5^4 = 25^2 )

\(=\frac{1}{25\times7}=\frac{1}{175}\)

\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)

\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)

\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)

\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)

a) \(\frac{-9}{12}=\frac{-9\div3}{12\div3}=\frac{-3}{4}\)

\(\frac{-18}{-24}=\frac{-18\div\left(-6\right)}{-24\div\left(-6\right)}=\frac{3}{4}\)

\(\frac{-35}{70}=\frac{-35\div35}{70\div35}=\frac{-1}{2}\)

\(\frac{-9}{27}=\frac{-9\div9}{27\div9}=\frac{-1}{3}\)

b) \(\frac{1313}{4242}=\frac{1313\div101}{4242\div101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{-353535\div\left(-70707\right)}{-424242\div\left(-70707\right)}=\frac{5}{6}\)

c)\(\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{\left(2^3\right)^2\times\left(5^2\right)^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^6\times5^6\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^3\times2^3\times5^2\times5^4\times7}\)= \(\frac{2^3\times2^3}{2^3\times5^2\times7}\)= \(\frac{8}{25\times7}\)= \(\frac{8}{175}\)

\(\left(2a+3\right)\left(2a+3\right)y+\left(2a+3\right)\)

\(=\left(2a+3\right)[y\left(2a+3\right)+1]\)

\(=\left(2a+3\right)\left(2ay+3y+1\right)\)

\(\left(a-b\right)x+\left(b-a\right)y-\left(a-b\right)\) (Sửa đề)

\(=\left(a-b\right)x-\left(a-b\right)y-\left(a-b\right)\)

\(=\left(a-b\right)\left(x-y-1\right)\)

\(\left(4x-y\right)\left(a+b\right)+\left(4x-y\right)\left(c-1\right)\)

\(=\left(4x-y\right)[\left(a+b\right)+\left(c-1\right)]\)

\(=\left(4x-y\right)\left(a+b+c-1\right)\)

Ta có a/(a+b+c)<a/(a+b)<a+c/a+b+c ( Cái này là vì a/a+b <1)

Tương tự vậy với mấy cái kia cx thế cộng theo vế là ra nha bạn 

Có ai giải rõ hơn k z ???

a,\(\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\Leftrightarrow\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{3x}{12}=\frac{2y}{4}=\frac{4z}{12}=\frac{3x-2y+4z}{12-4+12}=\frac{20}{20}=1\)

Suy ra:\(\hept{\begin{cases}\frac{x}{4}=1\\\frac{y}{2}=1\\\frac{z}{3}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=2\\z=3\end{cases}}\)

b, Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{2}=\frac{y}{6}=\frac{x-y}{2-6}=\frac{10}{-4}=-\frac{5}{2}\)

Suy ra:\(\hept{\begin{cases}\frac{x}{2}=-\frac{5}{2}\\\frac{y}{6}=-\frac{5}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-15\end{cases}}}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k;y=3k;z=4k\)

Do đó: \(xyz=2k.3k.4k=24\)

\(\Leftrightarrow24k=24\Rightarrow k=1\)

Từ k=1 ta tìm được x=2;y=3 và z=4