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AT
15 tháng 7 2021
a) \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{\left(2+\sqrt{a}-\sqrt{a}-1\right)\left(2+\sqrt{a}+\sqrt{a}+1\right)}{2\sqrt{a}+3}\)
\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)
b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
15 tháng 7 2021
a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)
\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)
Vậy ta có đpcm
b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)
Vậy ta có đpcm
7 tháng 10 2023
a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)
\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)
b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)
\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)
c:
\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)
d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
e:
\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)
f:
\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)
\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)
14 tháng 5 2023
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
8 tháng 2 2021
1) Biểu thức này là P hả?
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
P = \(\dfrac{\sqrt{a^3}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a^3}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\dfrac{a-1}{\sqrt{a}}\right).\left(\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\right)\)
= \(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\sqrt{a}}\)= \(\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)+2a+2}{\sqrt{a}}\)
= \(\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1+2a+2}{\sqrt{a}}\)
= \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
2) Để P = 7 với a ∈ ĐKXĐ
⇒ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) = 7
⇔ 2a + 2√a+2 = 7√a
⇔ 2a - 5√a + 2 = 0
⇔ \(\left[{}\begin{matrix}a=2\\a=\dfrac{1}{2}\end{matrix}\right.\)( thoả mãn ĐKXĐ)
Vậy...
3) Để P > 6 với a ∈ ĐKXĐ
⇒ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) >6
⇔ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) - 6 > 0
⇔ \(\dfrac{2a+2\sqrt{a}-6\sqrt{a}+2}{\sqrt{a}}>0\)
Mà √a > 0 với ∀a ∈ ĐKXĐ
⇒ 2a - 4√a + 2 >0
⇔ 2(√a - 1)2 > 0
Do 2(√a - 1)2 ≥ 0 với ∀a ∈ ĐKXĐ
Nên để 2(√a - 1)2 > 0 ⇔ 2(√a - 1)2 ≠ 0
⇔ a ≠ 1
Đối chiếu ĐKXĐ ta được: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
Vậy để P > 6 thì a ∈ ĐKXĐ
8 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
1) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\dfrac{a}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\left(\dfrac{a+2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
2) Để P=7 thì \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\)
\(\Leftrightarrow2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow2a+2\sqrt{a}-7\sqrt{a}+2=0\)
\(\Leftrightarrow2a-5\sqrt{a}+2=0\)
\(\Leftrightarrow2a-4\sqrt{a}-\sqrt{a}+2=0\)
\(\Leftrightarrow2\sqrt{a}\left(\sqrt{a}-2\right)-\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\2\sqrt{a}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\2\sqrt{a}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\\\sqrt{a}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\left(nhận\right)\\a=\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: Để P=7 thì \(a\in\left\{4;\dfrac{1}{4}\right\}\)
7 tháng 11 2021
\(P=5\sqrt{a}+7\sqrt{a}-8\sqrt{a}=4\sqrt{a}\\ Q=\left[2+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[2-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\\ Q=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a\)
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.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
GD
30 tháng 8 2017
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
\(ĐKXĐ:-1\le a\le1\)
Đặt \(A=\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\)
\(\Leftrightarrow A^2=a+2\sqrt{a-1}+a-2\sqrt{a-1}+2\sqrt{\left(a+2\sqrt{a-1}\right)\left(a-2\sqrt{a-1}\right)}\)
\(\Leftrightarrow A^2=2a+2\sqrt{a^2-4\left(a-1\right)}\)
\(\Leftrightarrow A^2=2a+2\sqrt{a^2-4a+4}\)
\(\Leftrightarrow A^2=2a+2\sqrt{\left(a-2\right)^2}\)
\(\Leftrightarrow A^2=2a+2\left(a-2\right)\)
\(\Leftrightarrow A^2=2a+2a-4\)
\(\Leftrightarrow A^2=4a-4\)
\(\Leftrightarrow A=2\sqrt{a-1}\)