Bài 1 : Rút gọn biểu thức với giả thiết các biểu thức đều có nghĩa
a) A = \(4\sqrt{\frac{25x}{4}}-\frac{8}{3}\sqrt{\frac{9x}{4}}-\frac{4}{3x}\sqrt{\frac{9x^3}{54}}\left(x>0\right)\)
b) B = \(\frac{x}{2}+\frac{3}{4}\sqrt{1-4x+4x^2}-\frac{3}{2}\left(x\le\frac{1}{2}\right)\)
Bài 3 : Giải PT
a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
b) \(\sqrt{4x^2-9}=2\sqrt{2x+3}\)
c) \(3x-7\sqrt{x}+4=0\)
Bài 4 : Trục căn thức mẫu và rút gọn
a) \(\frac{9}{\sqrt{3}}\)
b) \(\frac{3}{\sqrt{5}-\sqrt{2}}\)
c) \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
d) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
Vậy thoiiiii :))) Giúp em với mọi người :")))
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)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
b)\(\sqrt{4x^2-9}=2\sqrt{2x+3}\) \(đk:x\ge-\frac{3}{2}\)
\(\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\sqrt{2x+3}=0\)
\(\sqrt{\left(2x+3\right)}\cdot\left(\sqrt{2x-3}-2\right)=0\)
\(\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{matrix}\right.\left[{}\begin{matrix}2x+3=0\\\sqrt{2x-3}=2\end{matrix}\right.\left[{}\begin{matrix}x=-\frac{3}{2}\\2x-3=4\left(tm\right)\\2x-3=-4\left(ktm\right)\end{matrix}\right.\left[{}\begin{matrix}x=-\frac{3}{2}\left(tm\right)\\x=\frac{7}{2}\left(tm\right)\end{matrix}\right.\)
c)\(3x-7\sqrt{x}+4=0\) \(đk:x\ge0\)
\(3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\left[{}\begin{matrix}3\sqrt{x}-4=0\\\sqrt{x}-1=0\end{matrix}\right.\left[{}\begin{matrix}3\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\left[{}\begin{matrix}\sqrt{x}=\frac{4}{3}\\x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\left[{}\begin{matrix}x=\frac{16}{9}\left(tm\right)\\x=-\frac{16}{9}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
bạn tự kết luận nha
bài 4
a \(\frac{9}{\sqrt{3}}\) =\(\frac{9\sqrt{3}}{3}\) = 3\(\sqrt{3}\)
b \(\frac{3}{\sqrt{5}-\sqrt{2}}\) =\(\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}\) = \(\left(\sqrt{5}+\sqrt{2}\right)\)
c \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\) = \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{2-1}\) = 1
d \(\frac{1}{7-4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\) =\(\frac{14}{5}\)
câu d mình giải tắt vì cách giải của nó cũng giống mấy câu trên
