Rút gọn : \(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\) với a≥0,a≠9
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4 tháng 2 2022
\(B=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}=\dfrac{11}{a-9}\)
26 tháng 7 2021
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
29 tháng 6 2023
a:
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
b: Khi x=7-4căn 3 thì
\(A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
c: A=3
=>căn x-2=1
=>x=9(loại)
Y
29 tháng 6 2023
\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{x-9}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{9-x+x-9-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-2}{4\sqrt{x}-4-x}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(x-4\sqrt{x}+4\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)
\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn đề bài.
H
18 tháng 9 2023
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
H9
HT.Phong (9A5)
CTVHS
19 tháng 9 2023
b) Xét hiệu:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)
\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)
\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)
Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\)
\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)
Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)
(giúp cậu nó nha)
9 tháng 5 2022
Với a >= 0 ; a khác 9
\(P=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{a-9}=\dfrac{3a-9\sqrt{a}}{a-9}=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)
b, Để hàm số trêm là hàm bậc nhất khi a khác 0
Cho (d') : y = ax - 4 Để (d') cắt (d) khi a khác -3
Thay y = 5 vào (d) ta được <=> 5 = -3x + 2 <=> x = -1
(d) cắt (d') tại A(-1;5)
<=> 5 = -a - 4 <=> a = -9 (tm)
12 tháng 7 2021
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x-6\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-9\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
a) \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\)
\(=\dfrac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\dfrac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}\)
\(=\dfrac{3-\sqrt{a}}{1}-\dfrac{\sqrt{a}-3}{1}\)
\(=3-\sqrt{a}-\sqrt{a}+3\)
\(=-2\sqrt{a}+6\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}-\sqrt{b}}{1}-\dfrac{\sqrt{a}-\sqrt{b}}{1}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
Y
26 tháng 6 2023
\(a,\dfrac{9-a}{\sqrt{a}+3}-\dfrac{9-6\sqrt{a}+a}{\sqrt{a}-3}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{-\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-\dfrac{\left(3-\sqrt{a}\right)^2}{3-\sqrt{a}}\)
\(=-\left(\sqrt{a}-3\right)+\left(3-\sqrt{a}\right)\)
\(=-\sqrt{a}+3+3-\sqrt{a}\)
\(=6-2\sqrt{a}\)
\(b,\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\left(dkxd:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
Y
27 tháng 6 2023
\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)
\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)
\(=\dfrac{11}{a-9}\)
\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
27 tháng 6 2023
bạn ơi có phải \(x\sqrt{x}\) là \(\left(\sqrt{x}\right)^3\) đúng ko ạ
28 tháng 6 2021
`M=sqrt{(3a-1)^2}+2a-3`
`=|3a-1|+2a-3`
`=3a-1+2a-3(do \ a>=1/3)`
`=5a-4`
`N=sqrt{(4-a)^2}-a+5`
`=|4-a|-a+5`
`=a-4-a+5(do \ a>4)`
`=1`
`I=sqrt{(3-2a)^2}+2-7`
`=|3-2a|-5`
`=3-2a-5(do \ a<3/2)`
`=-2-2a`
`K=(a^2-9)/4*sqrt{4/(a-2)^2}`
`=(a^2-9)/4*|2/(a-2)|`
`=(a^2-9)/(2|a-2|)`
Nếu `3>a>2=>|a-2|=a-2`
`=>K=(a^2-9)/(2(a-2))`
Nếu `a<2=>|a-2|=2-a`
`=>K=(a^2-9)/(2(2-a))`
28 tháng 6 2021
\(M=\left|3a-1\right|+2a-3\)
Mà \(a-\dfrac{1}{3}\ge0\)
\(\Rightarrow M=3a-1+2a-3=5a-4\)
\(N=\left|4-a\right|-a+5\)
Mà \(4-a< 0\)
\(\Rightarrow N=a-4-a+5=1\)
\(I=\left|3-2a\right|-5\)
Mà \(a-\dfrac{3}{2}< 0\)
\(\Rightarrow I=3-2a-5=-2a-2\)
K, Ta có : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
\(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3a-9\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)
Anh giúp em ạ! Anh có cách nào không cần chia TH không ạ
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