p= \(\dfrac{2\sqrt{x}+|\sqrt{x}-1}{3x+2\sqrt{x}-1}\)
a. Rút gọn P
b. tính giá trị của P với x=\(\dfrac{4}{9}\); x=\(\dfrac{9}{4}\)
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15 tháng 8 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)
15 tháng 8 2021
a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3=3\)
hay x=0
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
22 tháng 11 2023
a: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{x-1}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}-2+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=9/4 thì \(P=\dfrac{3}{2}:\left(\dfrac{3}{2}-1\right)=\dfrac{3}{2}:\dfrac{1}{2}=3\)
c: P<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<=x<1
AT
10 tháng 7 2021
a) \(P=\dfrac{x^2+3x}{x^2-8x+16}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\right)\left(x\ne0,x\ne4\right)\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x\left(x-4\right)}\right)\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{\left(x+4\right)\left(x-4\right)+x+19-x^2}{x\left(x-4\right)}\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{x+3}{x\left(x-4\right)}=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\dfrac{x\left(x-4\right)}{x+3}=\dfrac{x^2}{x-4}\)
b) \(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
\(\Rightarrow P=\dfrac{2^2}{2-4}=-2\)
10 tháng 7 2021
a)\(ĐKXĐ:\left\{{}\begin{matrix}x\left(x-4\right)\ne0\\\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne0\\x\ne-3\end{matrix}\right.\)
\(P=\dfrac{x\left(x+3\right)}{\left(x-4\right)}:\left(\dfrac{x^2-16+x+19-x^2}{x\left(x-4\right)}\right)=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\left(\dfrac{x\left(x-4\right)}{x+3}\right)=\dfrac{x^2}{x-4}\)
b)\(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3+1}-\left(\sqrt{3}-1\right)=2\)
thay x=2 vào P ta có \(P=\dfrac{2^2}{2-4}=-2\)
16 tháng 10 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
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
NB
23 tháng 6 2021
a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)
\(\left(x\ge0;x\ne4;9\right)\)
b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)
Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)
T
8 tháng 11 2023
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-4}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) Với \(x\ge0;x\ne1;x\ne4\):
Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:
\(P=\dfrac{\sqrt{3+2\sqrt{2}}+2}{\sqrt{3+2\sqrt{2}}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+2}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}\)
\(=\dfrac{\sqrt{2}+3}{\sqrt{2}}\)
\(=\dfrac{2+3\sqrt{2}}{2}\)
c) Với \(x\ge0;x\ne1;x\ne4\),
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên
\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)
\(\Rightarrow x\in\left\{4;16;0\right\}\)
Kết hợp với ĐKXĐ của \(x\), ta được:
\(x\in\left\{0;16\right\}\)
Vậy: ...
\(\text{#}Toru\)
6 tháng 12 2021
a: \(P=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
xem lại chỗ dấu giá trị tuyệt đối (hoặc không có thì nói:v)
Sửa đề; \(P=\dfrac{2\sqrt{x}+\sqrt{x}-1}{3x+2\sqrt{x}-1}\)
a: \(P=\dfrac{3\sqrt{x}-1}{3x+3\sqrt{x}-\sqrt{x}-1}=\dfrac{\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)
b: Khi x=4/9 thì P=1/(2/3+1)=1:5/3=3/5
Khi x=9/4 thì P=1/(3/2+1)=1:5/2=2/5