\(P=\dfrac{1}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+1}+\dfrac{2\sqrt{x}-7}{x-\sqrt{x}-2} \) với x\(\ge\)0;x\(\ne\)44
a) CM \(P=\dfrac{1}{\sqrt{x}-2}
\)
b) tìm giá trị lớn nhất của P
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24 tháng 11 2022
\(M=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-1}{x-1}\)
\(=\dfrac{-2\sqrt{x}}{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{-2}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
22 tháng 11 2022
c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)
\(=\dfrac{2}{3-\sqrt{7}}+\sqrt{7}\)
\(=3+2\sqrt{7}\)
d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
28 tháng 8 2023
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
9 tháng 7 2022
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
8 tháng 4 2023
a: \(P=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+8}{\sqrt{x}+1}\)
b: Khi x=3-2 căn 2 thì \(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)
23 tháng 4 2021
Ta có: \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
13 tháng 6 2023
1: \(P=\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}=\dfrac{\sqrt{x}-1}{x+1}\)
2: P<1/2
=>P-1/2<0
=>\(2\sqrt{x}-2-x-1< 0\)
=>-x+2căn x-1<0
=>(căn x-1)^2>0(luôn đúng)
a: \(P=\dfrac{\sqrt{x}+1-2\sqrt{x}+4+2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)
b: căn x+1>=1
=>P<=1
Dấu = xảy ra khi x=0