Question

[Bếu Khá BảnH]

\(đk:x\ne4;x\ge0\)

\(A=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)

\(B=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)

a) tìm x để B=A+1

b)C=B-A tìm GTNN của C

Answer 1

[Hy Minh]

a) Để B = A + 1 thì:

\(\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}+1\)

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

\(\Leftrightarrow\frac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\frac{2x-2\sqrt{x}-4}{\sqrt{x}-2}\)

\(\Leftrightarrow x-1=\frac{2\left(x-\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(\Leftrightarrow x-1=\frac{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\)

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

\(\Leftrightarrow x-2\sqrt{x}-1-2=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{2}\right)\left(\sqrt{x}-1+\sqrt{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1+\sqrt{2}\\\sqrt{x}=1-\sqrt{2}\end{matrix}\right.\) ( Loại \(\sqrt{x}=1-\sqrt{2}\) vì \(\sqrt{x}\ge0\) )

Vậy \(x=3+2\sqrt{2}\)

b) Ta có: B = x -1 ( theo kết quả rút gọn ở câu a )

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

Do đó: \(C=B-A=x-1-2\sqrt{x}-1\)

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

\(C=\left(\sqrt{x}-1\right)^2-3\ge-3\) với mọi x

Dấu bằng xảy ra khi: \(\sqrt{x}-1=0\Rightarrow x=1\)

Vậy min C = -3 khi và chỉ khi x = 1

Answer 2

[₮ØⱤ₴₮]

b) đk: ...\(A=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)

\(B=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=x-1\)biết B=A-1=>\(x-1=2\sqrt{x}+1+1\) giải nốt ra đc nghiệm x=9

KL: vậy ...