\(\dfrac{12}{47}=\dfrac{372}{1457}\\ \dfrac{8}{31}=\dfrac{376}{1457}\\ 372< 376\\ =>\dfrac{12}{47}< \dfrac{8}{31}\)

\(c,5\sqrt{12x}-4\sqrt{3x}+2\sqrt{48x}=14\\ 5.2\sqrt{3x}-4\sqrt{3x}+2.4\sqrt{3x}=14\\ 14\sqrt{3x}=14\\ \sqrt{3x}=1\\ 3x=1\\ x=\dfrac{1}{3}\)

\(d,2\sqrt{x-1}+\dfrac{3}{2}\sqrt{4x-4}-\dfrac{1}{3}\sqrt{9x-9}=12\\ 2\sqrt{x-1}+\dfrac{3}{2}.2\sqrt{x-1}-\dfrac{1}{3}.3\sqrt{x-1}=12\\ 4\sqrt{x-1}=12\\ \sqrt{x-1}=3\\ x-1=9\\ x=10\)

\(e,\dfrac{2}{5}\sqrt{25x+25}-3\sqrt{x+1}+\dfrac{5}{2}\sqrt{4x+4}=8\\ \dfrac{2}{5}.5\sqrt{x+1}-3\sqrt{x+1}+\dfrac{5}{2}.2\sqrt{x+1}=8\\ 4\sqrt{x+1}=8\\ \sqrt{x+1}=2\\ x+1=4\\ x=3\)

\(g,3\sqrt{2x}+4\sqrt{8x}-\sqrt{18x}=10\\ 3\sqrt{2x}+4.2\sqrt{2x}-3\sqrt{2x}=10\\ 8\sqrt{2x}=10\\ \sqrt{2x}=\dfrac{5}{4}\\ 2x=\dfrac{25}{16}\\ x=\dfrac{25}{32}\)

\(5x^2??4x\)

8 mak :))???

\(\text{a},\text{a}=x^2+2x+y^2-2y-2xy+37=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37=\left(7\right)^2+2.7+37=100\\ b,B=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10=\left(x+2y\right)^2-2\left(x+2y\right)+10=\left(5\right)^2-2.5+10=25\)

\(x-\left[-x+x+3\right]-\left(x+3-x+2\right)=0\\ x-3-5=0\\ x=8\)

\(1,=16x^2+2.4x.3+9\\ 2=x^2+2.x.6y+36y^2\\ 3,=\left(x^2-1\right)\\ 4,=\left(x^2-\dfrac{4}{25}y^2\right)\\ 5,=9x^2-2.3x.4y+16y^2\\ 6,=4x^2-2.2x.5y+25y^2\\ 7,=4x^2-2.2x.\left(-\dfrac{1}{2}y\right)+\dfrac{1}{4}y^2\\ 8,=\left(\dfrac{x^2}{4}-4y^4\right)\\ 9,=\left(9x^4-16\right)\\ 10,=25x^2+2.5x.2y+4y^2+16x^2-2.4x.\left(-3y\right)+9y^2=41x^2+44xy+13y^2\)

cho góc xOy nhọn , tù hay vuông :))

\(c,=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)=2x^2+2x+13-2x^2+2=2x+15\\ d,=x^2-4x+4+x^3-1-x\left(x^2-4\right)=x^3+x^2-4x+3-x^3+4x=x^2+3\)