Question

Dạng 3 bài 1 câu b kiểu j v

Answer 1

Dạng 3 Bài 1)

`B=(sqrtx/(x-4)+1/(sqrtx-2)):(sqrtx+2)/(x-4)(x>=0,x ne 4)`

`=(sqrtx/(x-4)+(sqrtx+2)/(x-4)):1/(sqrtx-2)`

`=(2sqrtx+2)/(x-4)*(sqrtx-2)`

`=(2sqrtx+2)/(sqrtx+2)`

`b)C=A(B-2)=(sqrtx+2)/(sqrtx-2)*(2sqrtx+2-2sqrtx-4)/(sqrtx+2)`

`=-2/(sqrtx-2)`

Vì `x in ZZ=>sqrtx-2 in ZZ`

`=>-2 vdots sqrtx-2`

`=>sqrtx-2 in Ư(-2)={+-1,+-2}`

`=>sqrtx in {1,3,0,4}`

`=>x in {1,9,0,16}`

Answer 2

Bài 2: 

a) Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)

\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{5\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

b) Ta có: \(C=\left(AB+\dfrac{x-5}{\sqrt{x}-5}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\cdot\dfrac{\sqrt{x}+6}{\sqrt{x}-1}+\dfrac{x-5}{\sqrt{x}-5}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}}\)

\(=\dfrac{x-5+\sqrt{x}+6}{\sqrt{x}-5}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Ta có: \(C-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{3\sqrt{x}}{\sqrt{x}}\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow C>3\)

Answer 3

Bài 3: 

a) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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