Sửa đề:
$\sqrt{3-x}-\sqrt{27-9x}+\frac{5}{4}\sqrt{48-16x}=6$
Lời giải:
ĐKXĐ: $x\leq 3$
PT $\Leftrightarrow \sqrt{3-x}-\sqrt{9(3-x)}+\frac{5}{4}\sqrt{16(3-x)}=6$
$\Leftrightarrow \sqrt{3-x}-3\sqrt{3-x}+5\sqrt{3-x}=6$
$\Leftrightarrow (1-3+5)\sqrt{3-x}=6$
$\Leftrightarrow 3\sqrt{3-x}=6$
$\Leftrightarrow \sqrt{3-x}=2$
$\Leftrightarrow 3-x=4$
$\Leftrightarrow x=-1$ (tm)
1. rút gọn biểu thức
B=\(\dfrac{3}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}-\dfrac{5\sqrt{6}}{2}\)
rút gọn
A=\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
cho x<0, y<0. Biểu thức A= \(x\sqrt{16y^2}\)bằng
rút gọn không dùng máy tính \(A=\frac{3+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}\)
RUT GON:
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
Rút gọn biểu thức
A=[1+(11-√11):(1-√11)]×[(11+√11):(1+√11)+1]
Cho biểu thức: \(P=\left(\frac{x+2}{\sqrt{x}+1}-\sqrt{x}\right):\left(\frac{\sqrt{x}-4}{1-x}-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\frac{x\sqrt{x}-1}{x-1}-\frac{x+1}{\sqrt{x}+1}\)
rút gọn biểu thức
1 .
a)\(A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
b)\(B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)
c)C=\(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)
2 .
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b)\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c)C=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}}\)
d)D=(\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Toán 9
(\(\frac{6-2\sqrt{2}}{3-2\sqrt{2}}\)-\(\frac{5}{\sqrt{5}}\)):\(\frac{1}{2-\sqrt{5}}\)
