p=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
a. rút gọn p với a ≥0 và a≠1
b. tìm a để p =0
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Những câu hỏi liên quan
22 tháng 12 2020
Bài 1:
a) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
hay a>4
Kết hợp ĐKXĐ,ta được: a>4
Vậy: Để Q dương thì a>4
10 tháng 9 2023
a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)}{\sqrt{a}-1}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: P<1
=>P-1<0
=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
=>\(\dfrac{a+2}{\sqrt{a}-1}< 0\)
=>căn a-1<0
=>0<=a<1
c: Khi a=19-8căn 3=(4-căn 3)^2 thì \(P=\dfrac{19-8\sqrt{3}+4-\sqrt{3}+1}{4-\sqrt{3}-1}=\dfrac{24-9\sqrt{3}}{3-\sqrt{3}}=\dfrac{15-\sqrt{3}}{2}\)
4 tháng 3 2021
a) \(P=\dfrac{1-2\sqrt{a}+a}{1-\sqrt{a}}\cdot\dfrac{1+2\sqrt{a}+a}{1+\sqrt{a}}\) \(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\) \(=1-a\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Để P>0 \(\Leftrightarrow1-a>0\) \(\Leftrightarrow a< 1\)
Vậy \(0\le a< 1\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right]\)
\(A=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(A=\dfrac{a-1}{\sqrt{a}}\)
b) Ta có:
\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{3+2\sqrt{3}}{\sqrt{3}+1}\)
c) \(A< 0\) khi:
\(\dfrac{a-1}{\sqrt{a}}< 0\)
Mà: \(\sqrt{a}\ge0\forall x\) (xác định)
\(\Leftrightarrow a-1< 0\)
\(\Leftrightarrow a< 1\)
Kết hợp với đk:
\(0< a< 1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
\(C=\left(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}+\sqrt{a}\)
\(=\left(1-2\sqrt{a}+a\right).\dfrac{1}{\left(\sqrt{a}-1\right)^2}+\sqrt{a}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2}+\sqrt{a}=\sqrt{a}+1\)
Để \(C=3\Rightarrow\sqrt{a}+1=3\Rightarrow\sqrt{a}=2\Rightarrow a=4\)
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: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
12 tháng 2 2021
a) Ta có: \(Q=\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-1}{x-1}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)^2\)
\(=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}:\dfrac{\left(x-\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\left(x-\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}\)
3 tháng 4 2022
\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(a>0;a\ne1\right)\\ =\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{a-1}{\sqrt{a}}\\ =\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{4a^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}\)
a) \(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)
\(=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\)
\(=\left(a-1\right)^2=a^2-2a+1\)
b) \(P=0\Rightarrow\left(a-1\right)^2=0\Rightarrow a=1\)
a) Ta có: \(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\)
\(=\left(a-1\right)^2\)
b) Để P=0 thì a-1=0
hay a=1(loại)