Ừ hồi nãy làm sai mình sửa lại r á:") nó không hiện hả?

9. Với \(x\ne2\):

\(A=\dfrac{x+1}{x-2}=\dfrac{x-2+3}{x-2}=1+\dfrac{3}{x-2}\)

Để A nguyên thì \(\dfrac{3}{x-2}\) nguyên \(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{-1;1;3;5\right\}\)

10. Với \(x\ne-5:\)

\(B=\dfrac{2x-1}{x+5}=\dfrac{2x+10-11}{x+5}=2-\dfrac{11}{x+5}\)

Để B nguyên thì \(\dfrac{11}{x+5}\) nguyên \(\Rightarrow\left(x+5\right)\inƯ\left(11\right)=\left\{-1;1;11;-11\right\}\Rightarrow x\in\left\{-6;-4;6;-16\right\}\)

#h24cfs_20022023

b này y hệt mình=)) chỉ khác là b vẫn muốn kết bạn với mn còn mình thì mặc kệ luôn

ĐK: \(x\ge0,x\ne4\)

\(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}=\dfrac{-1}{2-\sqrt{x}}+\dfrac{5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{2+\sqrt{x}}{\sqrt{x}}+\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4-x+x}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}:\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(2-\sqrt[]{x}\right)}{4}=\sqrt{x}-1\)

ĐK: \(x\ge0,x\ne4\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{x-4}=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}:\dfrac{-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)^2}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}.\dfrac{\left(\sqrt{x}+2\right)^2}{-\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

Vậy...

a) \(\dfrac{39}{41}=\dfrac{39.71}{41.71}=\dfrac{2769}{2911},\dfrac{69}{71}=\dfrac{69.41}{71.41}=\dfrac{2829}{2911}\)

\(2769< 2829\Rightarrow\dfrac{2769}{2911}< \dfrac{2829}{2911}\Rightarrow\dfrac{39}{41}< \dfrac{69}{71}\)

b) \(\dfrac{59}{-58}=\dfrac{59.\left(-68\right)}{-58.\left(-68\right)}=\dfrac{-4012}{3944},\dfrac{-69}{68}=\dfrac{-69.58}{68.58}=\dfrac{-4002}{3944}\)

\(-4012< -4002\Rightarrow\dfrac{-4012}{3944}< \dfrac{-4002}{3944}\Rightarrow\dfrac{59}{-58}< \dfrac{-69}{68}\)

c) \(\dfrac{-49}{50}=\dfrac{-49.9}{50.9}=\dfrac{-441}{450},\dfrac{91}{-90}=\dfrac{91.\left(-5\right)}{-90.\left(-5\right)}=\dfrac{-455}{450}\)

\(-441>-455\Rightarrow\dfrac{-441}{450}>\dfrac{-455}{450}\Rightarrow\dfrac{-49}{50}>\dfrac{91}{-90}\)

d) \(\dfrac{1399}{-1411}=\dfrac{1399.\left(-1151\right)}{-1411.\left(-1151\right)}=\dfrac{-1610249}{1624061},\dfrac{1169}{-1151}=\dfrac{1169.\left(-1411\right)}{-1151\left(-1411\right)}=\dfrac{-1649459}{1624061}\)

\(-1610249>-1649459\Rightarrow\dfrac{-1610249}{1624061}>\dfrac{-1649459}{1624061}\Rightarrow\dfrac{1399}{-1411}>\dfrac{1169}{-1151}\)

a) \(M+\left(12x^2-6xy\right)=3x^2+4xy-2y^2\Rightarrow M=3x^2+4xy-2y^2-12x^2+6xy=-9x^2-2y^2+10xy\)

b) \(M-\left(5xy-6y^2\right)=x^2-87xy+6y^2\Rightarrow M=x^2-87xy+6y^2+5xy-6y^2=x^2-82xy\)

c) \(\left(-20x^2y+15xy-y^3\right)-M=21x^2y-2y^3\Rightarrow M=-20x^2y+15xy-y^3-21x^2y+2y^3=-41x^2y+15xy+y^3\)

d) \(M+\left(12x^4-15x^2y+2xy^2+7\right)=0\Rightarrow M=-12x^4+15x^2y-2xy^2-7\)