P= \(\frac{x+2}{x\sqrt{x}-1}+\frac{1}{1-\sqrt{x}}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
a tìm đkxđ và rút gọn P
b tìm x sao cho P = \(\frac{1}{7}\)
c tìm x sao cho P nguyên
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ĐKXĐ: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9
Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{3\left(\sqrt{x}+1\right)}{x-5\sqrt{x}+6}\)
\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-4+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{2}{\sqrt{x}-3}\)
b) Ta có: P < -1 <=> \(\frac{2}{\sqrt{x}-3}< -1\) <=> \(\frac{2}{\sqrt{x}-3}+1< 0\)
<=> \(\frac{2+\sqrt{x}-3}{\sqrt{x}-3}< 0\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}-3}< 0\)
TH1: \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-3>0\end{cases}}\) <=> \(\hept{\begin{cases}x< 1\\x>9\end{cases}}\)(loại)
TH2: \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-3< 0\end{cases}}\) <=> \(\hept{\begin{cases}x>1\\x< 9\end{cases}}\)
Kết hợp vs đk => S = {x|1 < x < 9 và x \(\ne\)4}
c) Để P nguyên <=> 2 \(⋮\)\(\sqrt{x}-3\) <=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Lập bảng: tự làm
@Edogawa Conan phân số thứ 2 bạn bị sai rồi \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)=x+2\sqrt{x}-3\)
trước phân số là dấu "-" phải đổi dấu
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(a,ĐKXĐ:x\ge0;x\ne1\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(P=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(P=\left(x+1\right)^2\left(x-1\right)^2\)
\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)
\(P=\left(x^2+x-x-1\right)^2\)
\(P=\left(x^2-1\right)^2\)
b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)
\(\left(2-\sqrt{3}\right)^2\)
\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)
\(x^2-1< 2-\sqrt{3}\)
\(x^2< 3-\sqrt{3}\)
\(x< \sqrt{3-\sqrt{3}}\)
a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có: \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)}-\sqrt{x}\right)\)
\(P=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2=\left(x-1\right)^2\)
b) Với x > = 0 và x khác 1
Ta có: \(P< 7-4\sqrt{3}\)
<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)
<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)
<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)
<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)
<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)