Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: \(-3\le x\le3;x\ne0\)
Đặt \(\sqrt{9-x^2}=a\left(a\ge0;a\ne3\right)\Rightarrow x^2=9-a^2\),khi đó pt đã cho trở thành:
\(\frac{9-a^2}{3+a}+\frac{1}{4\left(3-a\right)}=1\)
\(\Rightarrow3-a+\frac{1}{4\left(3-a\right)}=1\)
\(\Rightarrow\frac{4\cdot\left(3-a\right)^2+1}{4\left(3-a\right)}=1\Rightarrow4a^2-24a+37=12-4a\)
\(\Rightarrow4a^2-20a+25=0\Rightarrow\left(2a-5\right)^2=0\Rightarrow2a-5=0\)
\(\Rightarrow a=\frac{5}{2}\)(tm điều kiện),theo cách đặt ta có
\(\sqrt{9-x^2}=\frac{5}{2}\Rightarrow9-x^2=\frac{25}{4}\Rightarrow x^2=\frac{11}{4}\Rightarrow x=\frac{\sqrt{11}}{2}\)(TMĐKXĐ)
Vậy pt đã cho có nghiệm duy nhất là \(x=\frac{\sqrt{11}}{2}\)
B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
\(\sqrt{2x+1}-\sqrt{3x}=x-1\)
ĐK: \(x\ge0\)
\(\sqrt{2x+1}-\sqrt{3x}=3x-\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{2x+1}-\sqrt{3x}=\left(\sqrt{3x}-\sqrt{2x+1}\right)\left(\sqrt{3x}+\sqrt{2x+1}\right)\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{3x}\right)\left(1+\sqrt{3x}+\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{3x}\Rightarrow x=1\left(tm\right)\)
Giair phương trình \(\frac{x+3}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}\)
= $\frac{3+x}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}$3+x3x =√19 +1x √49 +2x2