Rút gọn biểu thức \(\frac{3}{\sqrt{7}-2}-\frac{14}{\sqrt{7}}+\sqrt{\left(\sqrt{7}-2\right)^2}\)
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
THỰC HIỆN PHÉP TÌNH VÀ RÚT GỌN CÁC BIỂU THỨC:
A=\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\)\(\sqrt{6}\)
B= \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right)\)\(:\frac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn các biểu thức:
\(a,\sqrt{\sqrt{3}+2}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\)
\(b,\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}\)
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
Rút gọn biểu thức
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
Rút gọn biểu thức \(B=\left(\frac{\sqrt{a-2}+2}{3}\right)\left(\frac{\sqrt{a-2}}{3+\sqrt{a-2}}+\frac{a+7}{11-a}\right):\left(\frac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\frac{1}{\sqrt{a-2}}\right)\)
Rút gọn các biểu thức sau
a)\(\left(\sqrt{14}+\sqrt{16}\right)\sqrt{5-\sqrt{21}}\)
b)\(\frac{5}{\sqrt{21}-4}+\frac{3\sqrt{7}-7\sqrt{3}}{\sqrt{7}-\sqrt{3}}\)
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
rút gọn biểu thức \(A=\frac{\sqrt{20}+2}{\sqrt{3}-1}-\frac{\sqrt{112}+4}{\sqrt{5}+1}+\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)\)
Rút gọn biểu thức:
\(Q=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x-4}\right)\)
đkxđ: \(x\ge0;x\ne4\)
\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)
\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)
\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)
đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
Rút gọn biểu thức
a)\(\frac{3\sqrt{7}+7\sqrt{3}}{\sqrt{21}}\)
b)\(\frac{\sqrt{2\left(\sqrt{2-\sqrt{7}}\right)^2}}{\sqrt{56}-4}\)
c)\(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right):\left(15-2\sqrt{6}\right)\)