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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)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
-\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x\right|-6x+9\)
\(x< 0\)
\(--->x+3-x-6x+9\)
\(=\left(x-x\right)-6x+3+9\)
\(=-6x+\left(3+9\right)=-6x+12\)
\(x>0\)
\(--->3+x+x-6x+9\)
\(=\left(x+x-6x\right)+\left(3+9\right)\)
\(=\left(2x-6x\right)+12\)
\(=4x+12\)
Bài 1:
a) 4x-\(\sqrt{9x^2-12x+4}\)
= 4x-\(\sqrt{\left(3x-2\right)^2}\)
= 4x-\(|3x-2|\)
= 4x-3x+2
= x+2
b) Thay x=\(\dfrac{2}{7}\)vào biểu thức A, ta có:
A= \(\dfrac{2}{7}+\dfrac{1}{2}\)= \(\dfrac{11}{14}\)
Bài 2:
a) \(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow\)\(\left(\sqrt{x^2+2x+1}\right)^2=\left(\sqrt{x+1}\right)^2\)
\(\Leftrightarrow\)x2+2x+1=x+1
\(\Leftrightarrow\)x2+2x+1-x-1=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)