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Ta có: \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=2+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
ĐKXĐ: \(-1\le a< 1\); \(a\ne0\)
\(P=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}^2}{\sqrt{\left(1-a\right)\left(1+a\right)}-\sqrt{1-a}^2}\right).\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
\(=\dfrac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{1+a-\left(1-a\right)}.\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
\(=\left(\dfrac{1+\sqrt{\left(1-a\right)\left(1+a\right)}}{a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)
- Với \(a>0\)
\(\Rightarrow P=\dfrac{\left(\sqrt{1-a^2}+1\right)\left(\sqrt{1-a^2}-1\right)}{a^2}=\dfrac{1-a^2-1}{a^2}=-1\)
- Với \(a< 0\)
\(\Rightarrow P=-\dfrac{\left(1+\sqrt{1-a^2}\right)^2}{a^2}\)
\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)
a) B xác định khi và chỉ khi :
\(\left\{{}\begin{matrix}a\ge0\\\sqrt[]{a}\ne0\\\sqrt[]{a}-1\ne0\\\sqrt[]{a}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)
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
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 \(a\in\left\{4;\dfrac{1}{4}\right\}\)
\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(a>0;a\ne1\right)\\ =\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{a-1}{\sqrt{a}}\\ =\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{4a^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}\)
A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)
=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)
Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
a , ĐKXĐ : bạn tự làm nhé
Đặt \(\sqrt{a}=x\) , khi đó biểu thức trỏ thành:
P = \(\dfrac{x^2.x-1}{x^2-x}-\dfrac{x^2.x+1}{x^2+x}+\left(x-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right)\)
= \(\dfrac{x^3-1}{x\left(x-1\right)}-\dfrac{x^3+1}{x\left(x+1\right)}+\left(\dfrac{x^2-1}{x}\right)\left(\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{x^2-1}\right)\)
= \(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)x}-\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)x}+\dfrac{\left(x^2+2x+1\right)+\left(x^2-2x+1\right)}{x}\)
= \(\dfrac{x^2+x+1+x^2-x+1}{x}+\dfrac{2x^2+2}{x}\)
= \(\dfrac{4x^2+4}{x}\)= \(\dfrac{4a+4}{\sqrt{a}}\)
Mấy câu sau dễ rồi
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