Question

Cho M = \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}-1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\) với x > 0, x khác 1

a, Rút gọn biểu thức M

b, Tìm x để M= \(\dfrac{9}{2}\)

c, So sánh M và 4

Answer 1

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

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

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

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

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

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

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

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

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

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

\(=\dfrac{5\sqrt{x}+2x+2}{\sqrt{x}}\)

xin lỗi bn nhé, mk chỉ bt làm đến đây thôi, k bt có đúng k

Answer 2

nếu đề này trong sách vnen 9 trang 28 thì bạn ghi sai đề nhé