1,Rút gọn biểu thức sau bằng cách đặt nhân tử chung:
M = ( 1+\(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\)).(1+\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\))
K
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5 tháng 5 2021
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
5 tháng 5 2021
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
28 tháng 7 2023
A) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\)
\(\Leftrightarrow\dfrac{3}{2}\sqrt{x-1}=6\)
\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)
Vậy, x=17
28 tháng 7 2023
A: \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
=>5/2*căn x-1-căn x-1=6
=>3/2*căn x-1=6
=>căn x-1=4
=>x-1=16
=>x=17
B:
a: ĐKXĐ: x>=0; x<>1
b: Sửa đề: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)
=căn x-1+x-căn x+1
=x
TA
17 tháng 7 2021
ĐK: `a \ne 1 ; a>=0`
`C=(\sqrta/(\sqrta-1)-\sqrta/(a-1)) : (\sqrta+1)/(a-1)`
`=(\sqrta/(\sqrta-1) - \sqrta/((\sqrta-1)(\sqrta+1))) . (a-1)/(\sqrta+1)`
`= (\sqrta(\sqrta+1)-\sqrta)/((\sqrta-1)(\sqrta+1)) . ((\sqrta-1)(\sqrta+1))/(\sqrta+1)`
`= a/(\sqrta+1)`
17 tháng 7 2021
Ta có: \(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-1}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\dfrac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{a}{\sqrt{a}+1}\)
30 tháng 6 2021
a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)
\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)
\(=-7\sqrt{3}\cdot\sqrt{xy}\)
b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
30 tháng 6 2021
$a)-7xy.\sqrt{\dfrac{3}{xy}}$
$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$
$=-7.\sqrt{\dfrac{xy}{3}}$
$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$
$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$
$=(\sqrt{a}+1)(b\sqrt{a}+1)$
26 tháng 9 2021
a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)
b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
26 tháng 9 2021
a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)
b: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
3 tháng 10 2021
\(a,M=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\\ M=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,M< \dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}< 0\\ \Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}< 0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}< 0\\ \Leftrightarrow\sqrt{a}-4< 0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a< 16\\ \Leftrightarrow0< a< 16\left(kết.hợp.ĐKXĐ\right)\)
18 tháng 8 2021
a: Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
18 tháng 8 2021
b: Ta có: \(\left(\sqrt{x}+1\right)\cdot A=x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\cdot\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=x\)
\(\Leftrightarrow x-2\sqrt{x}+1=0\)
\(\Leftrightarrow x=1\left(loại\right)\)
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
18 tháng 5 2021
`a)đk:a>0,a ne 9`
`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`
`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`
`=2/(sqrta+3)`
`b)A>1/2`
`<=>2/(sqrta+3)>1/2`
`<=>sqrta+3<4`
`<=>sqrta<1`
`<=>a<1`
KẾt hợp đkxđ:`0<x<1`
18 tháng 5 2021
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)
\(=\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{2}{\sqrt{a}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)
mà \(2\left(\sqrt{a}+3\right)>0\forall a\)
nên \(1-\sqrt{a}>0\)
\(\Leftrightarrow\sqrt{a}< 1\)
hay a<1
Kết hợp ĐKXĐ, ta được: 0<a<1
đk a ≠ 1 ; a ≥ 0
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1+\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)