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
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 ...
B=A+1\(\Leftrightarrow\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{2x-3\sqrt{x}-2+\sqrt{x}-2}{\sqrt{x}-2}\)
\(\Leftrightarrow\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{2x-2\sqrt{x}-4}{\sqrt{x}-2}\)
\(\Leftrightarrow\left(\sqrt{x^3}-\sqrt{x}+2x-2\right)\left(\sqrt{x}-2\right)=\left(2x-2\sqrt{x}-4\right)\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow x^2-2\sqrt{x^3}-5x+2x\sqrt{x}+4=2x\sqrt{x}+2x-8\sqrt{x}-8\)
\(\Leftrightarrow x^2-2\sqrt{x^3}-3x+8\sqrt{x}+12=0\)
tìm x nha..Mk mới lớp 8 chưa đủ trình
