Ta có:

\(3x-3=3\left(x-1\right)\)

\(4-4x=-4\left(x-1\right)\)

\(x^2-1=\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow\) MTC là \(3.\left(-4\right).\left(x-1\right)\left(x+1\right)=-12\left(x-1\right)\left(x+1\right)\)

Do đó:

\(\dfrac{11x}{3x-3}=\dfrac{11x}{3\left(x-1\right)}=\dfrac{11x.\left(-4\right).\left(x+1\right)}{3\left(x-1\right).\left(-4\right)\left(x+1\right)}=\dfrac{-44x\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{5}{4-4x}=\dfrac{5}{-4\left(x-1\right)}=\dfrac{5.3\left(x+1\right)}{-4\left(x-1\right).3\left(x+1\right)}=\dfrac{15\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2x}{x^2-1}=\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x.\left(-12\right)}{-12\left(x-1\right)\left(x+1\right)}=\dfrac{-24x}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{-3}{16};\dfrac{-21}{56}\) MSC : 112

\(\dfrac{-3}{16}=\dfrac{\left(-3\right)\cdot7}{16\cdot7}=\dfrac{-21}{112}\)

\(\dfrac{-21}{56}=\dfrac{\left(-21\right)\cdot2}{56\cdot2}=\dfrac{-42}{112}\)

Vậy \(\dfrac{-3}{16};\dfrac{-21}{56}\) quy đồng lên ta được 2 phân số : \(\dfrac{-21}{112};\dfrac{-42}{112}\)

75/1225

-558/1225

\(\dfrac{3}{49};\dfrac{-12}{25}\)

\(\dfrac{3}{49}=\dfrac{3\cdot25}{49\cdot25}=\dfrac{75}{1225}\)

\(\dfrac{-12}{25}=\dfrac{\left(-12\cdot49\right)}{25\cdot49}=\dfrac{-558}{1225}\)

\(-\dfrac{12}{25}=\dfrac{-12.49}{25.49}=-\dfrac{558}{1225}\)

\(\dfrac{3}{49}=\dfrac{3.25}{49.25}=\dfrac{75}{1225}\)

làm cho rồi đó : https://hoc24.vn/cau-hoi/c-316-21-phan-56-quy-dong-a.5608508996971

\(-\dfrac{3}{16}=\dfrac{-3.7}{16.7}=-\dfrac{21}{112}\)

\(-\dfrac{21}{56}=\dfrac{-21.2}{56.2}=-\dfrac{42}{112}\)

Câu c mình làm rồi: Mn ơi, hướng dẫn em cách để giống mẫu đi ạ! - Hoc24

\(d,\dfrac{x}{x^3-27}=\dfrac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{x+2}{x^2-6x+9}=\dfrac{x+2}{\left(x-3\right)^2}=\dfrac{\left(x+2\right)\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{x-1}{x^2+3x+9}=\dfrac{\left(x-1\right)\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(f,\dfrac{x+2}{x^2-3x+2}=\dfrac{x+2}{\left(x-1\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(x-1\right)\left(x-2\right)\left(2x-3\right)}\\ \dfrac{x}{-2x^2+5x-3}=\dfrac{-x}{\left(2x-3\right)\left(x-1\right)}=\dfrac{-x\left(x-2\right)}{\left(2x-3\right)\left(x-1\right)\left(x-2\right)}\\ \dfrac{2x+1}{-2x^2+7x-6}=\dfrac{-\left(2x+1\right)}{\left(x-2\right)\left(2x-3\right)}=\dfrac{-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)\left(2x-3\right)}\)

\(\dfrac{a+x}{6x^2-ax-2a^2}=\dfrac{\left(a+x\right)}{\left(2x+a\right)\left(3x-2a\right)}\)

\(\dfrac{a-x}{3x^2+4ax-4a^2}=\dfrac{a-x}{\left(x+2a\right)\left(3x-2a\right)}\)

Do đó ta quy đồng:

\(\dfrac{a+x}{6x^2-ax-2a^2}=\dfrac{\left(a+x\right)\left(x+2a\right)}{\left(x+2a\right)\left(2x+a\right)\left(3x-2a\right)}\)

\(\dfrac{a-x}{3x^2+4ax-4a^2}=\dfrac{\left(a-x\right)\left(2x+a\right)}{\left(x+2a\right)\left(2x+a\right)\left(3x-2a\right)}\)

\(\dfrac{2}{2x+1}-1\ge0\Leftrightarrow\dfrac{2}{2x+1}-\dfrac{2x+1}{2x+1}\ge0\)

\(\Leftrightarrow\dfrac{2-\left(2x+1\right)}{2x+1}\ge0\)

\(\Leftrightarrow\dfrac{1-2x}{2x+1}\ge0\)

\(\Rightarrow-\dfrac{1}{2}< x\le\dfrac{1}{2}\)

a: 27/-180=-27/180=-3/20=-21/140

-6/-35=6/35=24/120

-3/-28=3/28=15/140

b: \(\dfrac{3\cdot4+3\cdot7}{6\cdot5+9}=\dfrac{3\left(4+7\right)}{30+9}=\dfrac{11}{13}=\dfrac{2849}{13\cdot259}\)

\(\dfrac{6\cdot9-2\cdot17}{63\cdot6-119}=\dfrac{54-34}{259}=\dfrac{20}{259}=\dfrac{260}{259\cdot13}\)

\(\dfrac{9}{8}=\dfrac{9.13}{8.13}=\dfrac{117}{104}\)

\(\dfrac{12}{13}=\dfrac{12.8}{13.8}=\dfrac{96}{104}\)

\(MSC:104\)

\(\dfrac{117}{104};\dfrac{96}{104}\)

Hai đường thẳng đồng quy tại một điểm có nghĩa là hai đường thẳng cắt nhau tại một điểm

là 2 đường thẳng cắt nhau tại 1 điểm