Cho biếu thức: P= \(\dfrac{a+4\sqrt{a}+4}{\sqrt{2+a}}+\dfrac{4-a}{2-\sqrt{a}}\) ( với a>o; a \(\ne\)4)
a) rút gọn biểu thức P.
b) tìm giá trị của a sao cho P=a+1.
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ND
23 tháng 6 2018
a. \(A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
\(A=\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(A=\dfrac{5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}}{5-4}-\dfrac{2+\sqrt{3}}{4-3}\)
\(A=\sqrt{3}-2-\sqrt{3}=-2\)
b.
\(B=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(B=\left[\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right].\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right).\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\dfrac{-8\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}\)
\(B=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
23 tháng 6 2018
\(a.A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\sqrt{3}-\dfrac{1}{2-\sqrt{3}}=\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}\)
\(b.\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\left(a>0;a\ne4\right)\)
10 tháng 8 2018
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
10 tháng 8 2018
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
B
15 tháng 5 2021
Bài làm :
1) Khi x=9 ; giá trị của A là :
\(A=\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)
2) Ta có :
\(B=...\)
\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
3) Ta có :
\(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+2}\div\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)
Xét :
\(\frac{A}{B}+1=\frac{4}{\sqrt{x+2}}>0\Rightarrow\frac{A}{B}>-1\)
=> Điều phải chứng minh
4 tháng 6 2021
1, thay x=9(TMĐKXĐ) vào A ta đk:
A=\(\dfrac{\sqrt{9}}{\sqrt{9}-2}=3\)
vậy khi x=9 thì A =3
2,với x>0,x≠4 ta đk:
B=\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
3,\(\dfrac{A}{B}>-1\) (x>0,x≠4)
⇒\(\dfrac{\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}>-1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}>-1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>-1\)
⇒\(\sqrt{x}-2>-1\) (vì \(\sqrt{x}+2>0\))
⇔\(\sqrt{x}>1\)⇔x=1 (TM)
vậy x=1 thì \(\dfrac{A}{B}>-1\) với x>0 và x≠4
25 tháng 11 2018
a) \(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{4a+2\sqrt{a}-4}{4-a}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{2\sqrt{a}-a}\right)=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\left[\dfrac{-2\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right]=\left[\dfrac{a+4\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{2\sqrt{a}-a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{a+4\sqrt{a}+4+2\sqrt{a}-a-4a-2\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{2-\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{-4a+4\sqrt{a}+8}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{1}{\sqrt{a}}=\dfrac{4\left(2-\sqrt{a}\right)\left(\sqrt{a}+1\right).\sqrt{a}}{\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)}=\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}\)
Ta có A=\(\sqrt{a}+2\Leftrightarrow\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}=\sqrt{a}+2\Leftrightarrow4a+4\sqrt{a}=\left(\sqrt{a}+2\right)^2\Leftrightarrow4a+4\sqrt{a}=a+4\sqrt{a}+4\Leftrightarrow3a=4\Leftrightarrow a=\dfrac{4}{3}\left(tm\right)\)Vậy a=\(\dfrac{4}{3}\) thì A=\(\sqrt{a}+2\)
HC
25 tháng 11 2018
\(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{2+\sqrt{a}}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)\(=\left(\dfrac{\left(2+\sqrt{a}\right)^2+\sqrt{a}\left(2-\sqrt{a}\right)-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\right)\)\(:\left(\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right)\)
\(=\dfrac{4+4\sqrt{a}+a+2\sqrt{a}-a-4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) . \(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\dfrac{-4a+8\sqrt{a}}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) .\(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}.\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
=\(\dfrac{4a}{2+\sqrt{a}}\)
b, Để A=\(\sqrt{a}+2\)
<=> \(\dfrac{4a}{2+\sqrt{a}}\) =\(\sqrt{a}+2\)
<=> 4a=\(\left(\sqrt{a}+2\right)^2\)
<=> \(a+4\sqrt{a}+4-4a=0\)
<=> \(-3a+4\sqrt{a}+4=0\)
<=>\(-3a+6\sqrt{a}-2\sqrt{a}+4=0\)
<=> \(-3\sqrt{a}\left(\sqrt{a}-2\right)-2\left(\sqrt{a}-2\right)=0\)
<=> \(\left(\sqrt{a}-2\right)\left(-3\sqrt{a}-2\right)=0\)
<=>\(\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{-2}{3}\left(vl\right)\end{matrix}\right.\)
<=> a=4
18 tháng 5 2021
`P=(sqrta+3)/(sqrta-2)-(sqrta-1)/(sqrta+2)+(4sqrta-4)/(4-a)`
`đk:x>=0,x ne 4`
`P=(a+5sqrta+6-a+3sqrta-2-4sqrta+4)/(a-4)`
`=(4sqrta+8)/(a-4)`
`=4/(sqrta-2)`
`b)a=9`
`=>P=4/(3-2)=4`
18 tháng 5 2021
a) Ta có: \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
\(=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\dfrac{4}{\sqrt{a}-2}\)
b) Thay a=9 vào P, ta được:
\(P=\dfrac{4}{\sqrt{9}-2}=\dfrac{4}{3-2}=\dfrac{4}{1}=4\)
Vậy: khi a=9 thì P=4
28 tháng 11 2022
a: \(=\dfrac{-a-2\sqrt{a}+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(=\dfrac{-4a-6\sqrt{a}+4}{a-4}\cdot\dfrac{-\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+2}\)
\(=\dfrac{4a+6\sqrt{a}-4}{\sqrt{a}+2}\cdot\dfrac{\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(4a+6\sqrt{a}-4\right)}{4-a}\)
b: Để \(A=\sqrt{a}+2\) thì \(4a\sqrt{a}+6a-4\sqrt{a}=\left(\sqrt{a}+2\right)\left(4-a\right)=4\sqrt{a}-a\sqrt{a}+8-2a\)
=>\(5a\sqrt{a}+8a-8\sqrt{a}-8=0\)
=>\(5a\cdot\sqrt{a}+10a-2a-4\sqrt{a}-4\sqrt{a}-8=0\)
=>\(\left(\sqrt{a}+2\right)\left(5a-2\sqrt{a}-4\right)=0\)
=>\(5a-2\sqrt{a}-4=0\)
=>\(a=\dfrac{22+2\sqrt{21}}{25}\)
13 tháng 7 2018
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)