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\(\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\\ =\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)\cdot\left(3\sqrt{x}+14\right)}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+14}{5\sqrt{x}-1}\)
a) ĐKXĐ: x \(\ge\)0; x \(\ne\)4
Ta có: P = \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\frac{x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-\left(x+6\sqrt{x}+\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
P = \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}\)
b) Với x \(\ge\)0 và x \(\ne\)4, ta có:
P > -1 <=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}>-1\)
<=> \(-\frac{\sqrt{x}+6}{\sqrt{x}-2}+1>0\)
<=> \(\frac{\sqrt{x}-2-\sqrt{x}-6}{\sqrt{x}-2}>0\)
<=> \(\frac{-8}{\sqrt{x}-2}>0\)
Do -8 < 0 => \(\sqrt{x}-2< 0\) <=> \(\sqrt{x}< 2\)<=> \(x< 4\)
mà x \(\ge0\) => 0 \(\le\)x \(< \)4
c)Với x \(\ge\)0 và x \(\ne\)4
Để P \(\in\)Z <=> -8 \(-8⋮\sqrt{x}-2\)
<=> \(\sqrt{x}-2\inƯ\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Do \(\sqrt{x}\ge0\) <=> \(\sqrt{x}-2\ge-2\) => \(\sqrt{x}-2\in\left\{-2;-1;1;2;4;8\right\}\)
Lập bảng:
| \(\sqrt{x}-2\) | -2 | -1 | 1 | 2 | 4 | 8 |
| x | 0 | 1 | 9 | 16 | 36 | 100 |
Vậy ....
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
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