\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-...">
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21 tháng 10 2021

\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

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

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

11 tháng 7 2018

\(a.P=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}=\dfrac{\sqrt{x}-3-5+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-8+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\) ( x ≥ 0 ; x # 9 )

\(b.\) \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-\sqrt{x}}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}}{\sqrt{x}+2}\text{≤}2\)

\(P_{Max}=2."="\)\(x=0\)

\(=\dfrac{8-x}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}\cdot\dfrac{\sqrt[3]{x^2}-1}{\sqrt[3]{x}\left(\sqrt[3]{x}+1\right)}\)

\(=2-\sqrt[3]{x}+\dfrac{\sqrt[3]{x}-1}{\sqrt[3]{x}-2}\)

\(=\dfrac{4-4\sqrt[3]{x}+\sqrt[3]{x^2}-\sqrt[3]{x}+1}{2-\sqrt[3]{x}}\)

\(=\dfrac{\sqrt[3]{x^2}-5\sqrt[3]{x}+5}{2-\sqrt[3]{x}}\)

15 tháng 11 2018

1, ĐKXĐ: x\(\ge0\);x\(\ne1\)

Rút gọn P với \(x\ge0;x\ne1\)ta có

P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)

=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)

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

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

15 tháng 11 2018

2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:

P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)

=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)

=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)

\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3

31 tháng 7 2018

hình như đề bài bị sai số thì phải bạn ạ

mình giải cứ bị lệch số ấy

Sửa đề: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{4-\sqrt{x}}\)

a: \(P=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{4-\sqrt{x}}=\dfrac{2x}{4-\sqrt{x}}\)

b: Để P>3 thì P-3>0

\(\Leftrightarrow-\dfrac{2x}{\sqrt{x}-4}-3>0\)

\(\Leftrightarrow\dfrac{-2x-3\sqrt{x}+12}{\sqrt{x}-4}>0\)

\(\Leftrightarrow\dfrac{5\sqrt{x}-12}{\sqrt{x}-4}< 0\)

=>12/5<căn x<4

=>144/25<x<16

22 tháng 5 2018

Mẫu thức chung là (√x+1)(√x−4)

Bạn quy đồng lên rồi tính là ra

P/s: mình hơi lười. Bạn thông cảm nhé

a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)

b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)

\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)

28 tháng 7 2018

Bạn ơi đề bài là jlolang

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

\(=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}-1}{x\sqrt{x}+1}\)

\(=\dfrac{2x^2+2\sqrt{x}-9x\sqrt{x}-9+2x\sqrt{x}-10x+12\sqrt{x}-x+5\sqrt{x}-6}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

\(=\dfrac{2x^2+19\sqrt{x}-7x\sqrt{x}-11x-15}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

b: \(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)

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

20 tháng 6 2017

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

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

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

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

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

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

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