rút gọn : \(\frac{1}{\sqrt{17-12\sqrt{2}}}+\frac{1}{\sqrt{90-70\sqrt{2}}}\)
rút gọn
\(E=\frac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}+\frac{3+2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\frac{1}{\sqrt{12+2\sqrt{35}}}\)
1) Rút gọn
h)\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)
e)\(\frac{\sqrt{28}-2\sqrt{12}-2\sqrt{18}}{3\sqrt{7}-2\sqrt{27}-\sqrt{102}}\)
f)\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
mọi ng giúp mình vs đang cần gấp tks !!!đc câu nào giúp câu đấy cx đk
\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)
a) Rút gọn biểu thức:
\(P=\frac{5+\sqrt{10}+\sqrt{17}}{2}\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-5\right)\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-\sqrt{10}\right)\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-\sqrt{17}\right).\)
b) Giải phương trình: \(\frac{x+2}{2x-1}+|\frac{4x-2}{x+2}|+1=0\)
1. Cho biểu thức: A=\(\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-x}{\sqrt{x}-1}\right)\left(1+\frac{1}{\sqrt{x}}\right)\)
a) Rút gọn biểu thức A
b) Tìm giá trị của x để A= 4
2. Rút gọn các biểu thức sau:
a) A= \(3\sqrt{12}-4\sqrt{3}+5\sqrt{27}\)
b) B= \(\frac{1}{\sqrt{7}+4\sqrt{3}}\)
3. Tính giá trị biểu thức D=\(\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)
Rút gọn A= \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{16}}\)
Ta có:
\(A=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{5}{12}-\frac{1}{16}}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{17}{48}}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{51}}{12}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{17}}{12}\)
\(A=\frac{4\sqrt{3}+2\sqrt{2}+\sqrt{17}}{12}\)
Ta có: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}=\sqrt{\frac{5-2\sqrt{6}}{12}}\)
Vì \(5-2\sqrt{6}=3-2\sqrt{3}.\sqrt{2}+2=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\)\(\Rightarrow5-2\sqrt{6}=\left(\sqrt{3}-\sqrt{2}\right)^2\)
Như vậy: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)
Lại có: \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}.\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)
Rút gọn ta được \(A=\frac{\sqrt{3}}{2}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5-2\sqrt{6}}{12}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}\)
\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}}\left(do\sqrt{3}-\sqrt{2}>0\right)\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{6}\left(\sqrt{3}-\sqrt{2}\right)=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}-\frac{\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
a,\(\frac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
b,\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)rút gọn
Rút gọn
\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\left(với\right)x\ge0,x\ne1\)
Tính
\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)
\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)
Bài làm
Rút gọn
\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}+1}{(\sqrt{x}-1)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Tính:
\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)
\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)
\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)
\(=6-7\sqrt{3}+7\sqrt{3}\)
\(=6\)
Bài làm
\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)
\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)
\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)
\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)
\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)
\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)
\(=5-4\)
\(=1\)
rút gọn
\(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(\frac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}-\frac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
Rút gọn \(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
\(=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\sqrt{3}+\frac{2+\sqrt{3}}{4-3}=2+2\sqrt{3}\)
P/s:Không chắc ạ,em mới lớp 7 thôi!