P=\(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a-1}}\right)\)
Rút gọn P
Tìm các gtri của a để P<0
Tìm các gtri của a để P=2
K
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Những câu hỏi liên quan
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 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
a) Ta có: \(P=\dfrac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\dfrac{6-2\left(\sqrt{a}-1\right)^2}{a\sqrt{a}-1}+\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{\sqrt{a}-1}{a+\sqrt{a}+1}-\dfrac{-2a+4\sqrt{a}+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{-2a+4\sqrt{a}+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{a-2\sqrt{a}+1+2a-4\sqrt{a}-4+2a+2\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5a-4\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5a-5\sqrt{a}+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(5\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5\sqrt{a}+1}{a+\sqrt{a}+1}\)
b) Để P=1 thì \(5\sqrt{a}+1=a+\sqrt{a}+1\)
\(\Leftrightarrow a+\sqrt{a}+1-5\sqrt{a}-1=0\)
\(\Leftrightarrow a-4\sqrt{a}=0\)
\(\Leftrightarrow\sqrt{a}\left(\sqrt{a}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\\\sqrt{a}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\left(nhận\right)\\a=16\left(nhận\right)\end{matrix}\right.\)
Vậy: Để P=1 thì \(a\in\left\{0;16\right\}\)
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
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
AH
Akai Haruma
Giáo viên
1 tháng 4 2021
Lời giải:
ĐK: $x>0; a\neq 1; a\neq 4$
a)
$M=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-2)(\sqrt{a}-1)}$
$=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}$
b)
$M>\frac{-1}{2}\Leftrightarrow \frac{\sqrt{a}-2}{3\sqrt{a}}+\frac{1}{2}>0$
$\Leftrightarrow \frac{5\sqrt{a}-4}{6\sqrt{a}}>0$
$\Leftrightarrow 5\sqrt{a}-4>0$
$\Leftrightarrow a>\frac{16}{25}$
Kết hợp với ĐKXĐ thì $a>\frac{16}{25}; a\neq 1; a\neq 4$
4 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{-4\sqrt{x}\cdot\left(x-1\right)}{4x}\)
\(=\dfrac{-x+1}{\sqrt{x}}\)
b) Để P=2 thì \(-x+1=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+2\sqrt{x}-1=0\)
\(\Leftrightarrow x+2\sqrt{x}+1-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=\sqrt{2}\\\sqrt{x}+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{2}-1\\\sqrt{x}=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3-2\sqrt{2}\)
Vậy: Để P=2 thì \(x=3-2\sqrt{2}\)
Y
25 tháng 5 2023
\(a,\) Rút gọn
\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)
\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)
\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)
\(=\dfrac{3+\left(3-\sqrt{7}\right)\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)
\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)
\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=5\)
Vậy \(A=5\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{x-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{x-\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}-1\)
Vậy \(B=\sqrt{x}-1\)
\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)
\(\Leftrightarrow\sqrt{x}< 6\)
\(\Leftrightarrow x< 36\)
a) DKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
P=\(\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\\ =\dfrac{\left(a-1\right)^2}{4a}.\left(\dfrac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
= \(\dfrac{a-1}{4a}.\dfrac{-2.2\sqrt{a}}{1}\)
= \(\dfrac{1-a}{\sqrt{a}}\)
b) P<0 với a ∈ DKXD
=> \(\dfrac{1-a}{\sqrt{a}}< 0\)
mà √a > 0 với ∀a ∈ DKXD
=> 1-a < 0
<=> a>1 ( thoả mãn DKXD)
Vậy để P<0 thì a>1.
c) Để P = 2 với a ∈ DKXD
=> \(\dfrac{1-a}{\sqrt{a}}=2\)
<=> 1-a = 2√a
<=> a + 2√a -1 = 0
<=> \(\left[{}\begin{matrix}\sqrt{a}=-1+\sqrt{2}\\\sqrt{a}=-1-\sqrt{2}\left(loại\right)\end{matrix}\right.\)
<=> a = \(\sqrt{\sqrt{2}-1}\)(thoả mãn DKXD)
Vậy để P =2 thì a = \(\sqrt{\sqrt{2}-1}\)
Sửa đề: \(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)^2}{4a}\cdot\dfrac{-4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)\cdot\left(-1\right)}{\sqrt{a}}\)
\(=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
\(=\dfrac{1-a}{\sqrt{a}}\)
b) Để P<0 thì \(\dfrac{1-a}{\sqrt{a}}< 0\)
mà \(\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ
nên 1-a<0
hay a>1
Kết hợp ĐKXĐ, ta được: a>1
Vậy: Để P<0 thì a>1
c) Để P=2 thì \(\dfrac{1-a}{\sqrt{a}}=2\)
\(\Leftrightarrow1-a=2\sqrt{a}\)
\(\Leftrightarrow2\sqrt{a}+a-1=0\)
\(\Leftrightarrow a+2\sqrt{a}+1-2=0\)
\(\Leftrightarrow\left(\sqrt{a}+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+1=\sqrt{2}\\\sqrt{a}+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=\sqrt{2}-1\\\sqrt{a}=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\)
hay \(a=3-2\sqrt{2}\)(nhận)
Vậy: Để P=2 thì \(a=3-2\sqrt{2}\)