oc cho

Bổ sung phần c và d luôn:

c, C = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) 5(x² - 1) = 2(2x² + 3)

\(\Leftrightarrow\) 5x² - 5 = 4x² + 6

\(\Leftrightarrow\) x² = 11

\(\Leftrightarrow\) x² - 11 = 0

\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)

d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)

C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x² + \(\dfrac{3}{2}\))

\(\Leftrightarrow\) 4(x² + \(\dfrac{3}{2}\)) \(\in\) Ư(5)

Xét các TH:

4(x² + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x² = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x² = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x² = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x² = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)

Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z

Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)

`B17:`

`a)` Với `x \ne +-3` có:

`A=[x+15]/[x^2-9]+2/[x+3]`

`A=[x+15+2(x-3)]/[(x-3)(x+3)]`

`A=[x+15+2x-6]/[(x-3)(x+3)]`

`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`

`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)

   `=>` Ko có gtr nào của `x` t/m

`c)A in ZZ<=>3/[x-3] in ZZ`

   `=>x-3 in Ư_3`

 Mà `Ư_3={+-1;+-3}`

`@x-3=1=>x=4`

`@x-3=-1=>x=2`

`@x-3=3=>x=6`

`@x-3=-3=>x=0`

________________________________

`B18:`

`a)M=1/3`             `ĐK: x  \ne +-4`

`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`

`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`

`<=>32/[x-4].[x+4]/32=1/3`

`<=>3x+12=x-4`

`<=>x=-8` (t/m)

a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)

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

c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)

\(\Rightarrow x\in\left\{0;1;9;16\right\}\)

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

\(C=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)

ĐKXĐ: \(x\ge0;x\ne4\)

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

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

b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)

c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)

\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)

a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b: Thay x=36 vào A, ta được:

\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

hay \(x=\dfrac{1}{4}\)

d: Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

e: Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{-1;1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;4;0\right\}\)

hay \(x\in\left\{1;9;16;0\right\}\)