\(\left(-7\right)\)2 . \(\frac{1}{99}\) - \(\sqrt{81}\) + \(-\left(4\right)\)4 : \(\sqrt{64}\) - \(\left|-27\right|\)
GIÚP MIK ZỚI CẢM ƠN NHÌU >.<
Giúp mik với
Tính
a)\(\frac{2}{3}\sqrt{81}-\left(\frac{-3}{4}\right).\sqrt{\frac{9}{64}}+\left(\frac{\sqrt{2}}{3}\right)^2\)
b)\(\left(-\sqrt{\frac{5}{4}}\right)^2-\sqrt{\frac{9}{4}}:\left(-4,5\right)-\sqrt{\frac{25}{16}}.\sqrt{\frac{64}{9}}\)
c)\(-2^4-\left(-2\right)^2:\left(-\sqrt{\frac{16}{121}}\right)-\left(-\sqrt{\frac{2}{3}}\right)^2:\left(-2\frac{2}{3}\right)\)
Tính
a) \(2\sqrt{\frac{25}{16}}-3\sqrt{\frac{49}{36}}+4\sqrt{\frac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\frac{1}{2}}\right)^2+\frac{1}{16}.\left(\sqrt{\frac{3}{4}}\right)^2\)
c) \(\frac{2}{3}\sqrt{\frac{81}{16}}-\frac{3}{4}\sqrt{\frac{64}{9}}+\frac{7}{5}.\sqrt{\frac{25}{196}}\)
a) = \(\frac{7}{2}\)
b) = \(\frac{643}{64}\)
c) = 0
\(Cho\)\(x=\left(1+\frac{1}{\sqrt{1}}\right)+\left(1+\frac{1}{\sqrt{9}}\right)+\left(\frac{1}{\sqrt{25}}\right)+\left(1+\frac{1}{\sqrt{49}}\right)+\left(1+\frac{1}{\sqrt{81}}\right)\)
\(y=\left(1+\frac{1}{\sqrt{4}}\right)+\left(1+\frac{1}{\sqrt{16}}\right)+\left(1+\frac{1}{\sqrt{36}}\right)+\left(1+\frac{1}{\sqrt{64}}\right)+\left(1+\frac{1}{\sqrt{100}}\right)\)
Tính x.y
Mn ơi, giúp mk nha, mai mk nộp òi!
=(1/1+1/3+1/5+1/7+1/9).(1/2+1/4+1/6+1/8+1/10)
=(1+1/3+1/5+1/7+1/9).137/120
=258/315.137/120
=5891/6300
Tính \(\frac{2}{3}\sqrt{81}-\left(-\frac{3}{4}\right):\sqrt{\frac{9}{64}}+\left(\frac{\sqrt{2}}{3}\right)^0-\left(\sqrt{3}\right)^2\)
Rút gọn rồi tính giá trị của biểu thức:
A= \(\sqrt{\frac{\left(x-6^{ }\right)^4}{\left(5-x\right)^2}}+\frac{x^2-36}{x-5}\left(x< 5\right)\)tại x = \(\sqrt{\frac{12}{5}}:\sqrt{\frac{48}{5}}.\sqrt{64}\)
B= 5x - \(\sqrt{125}\) + \(\frac{\sqrt{x^3+5x^2}}{\sqrt{x+5}}\left(x>=0\right)\)tại x = \(\sqrt{\frac{65}{17}}:\sqrt{\frac{13}{4}}\)
C= \(\sqrt{\frac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\frac{\sqrt{x^4-2x^2+1}}{x-3}\left(x< 3\right)\)tại x =\(\sqrt{\frac{1}{18}}:\frac{1}{\sqrt{81}}\)
Các bác giúp e vs ạ, hứa sẽ tick, e cảm ơn nhiều!!!!!!!!
\(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}.\)
\(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right):\sqrt{\frac{81}{6}}\)
AI ĐÓ TỐT BỤNG GIÚP MK ZỚI MAI MK KTRA RÙI!!!
Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)
\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)
\(=\frac{22-5\sqrt{30}}{6}\)
Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)
\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)
\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)
\(=-\frac{67}{54}\)
Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
Sao làm hổng ai bảo đú.n/g vậy :(((
Tính nhanh:
\(\frac{3-3^2+3^3-3^4+...+3^{99}}{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}}.\left(11-\sqrt{91}\right)\left(11-\sqrt{95}\right)\left(11+\sqrt{99}\right)\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)...\left(11-\sqrt{113}\right)\left(11-\sqrt{104}\right)\)
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu
Giải giúp mik vs đap cần gấp . Cảm ơn mn. Giải cho mik bài 1 cx đc
1/ Rút gọn
A=\(\sqrt{7}-4\sqrt{3}+\sqrt{4}-2\sqrt{3}\)
B=\(\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)\) \(\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
C=\(\left(\sqrt{3}+1\right)\) \(\left(\frac{\sqrt{14}-6\sqrt{3}}{5+\sqrt{3}}\right)\)
2/Cho P=\(\left(\sqrt{x-\frac{1}{\sqrt{x}}}\right)\):\(\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
a/ cmr: P>0, V x >0, x\(\ne\)1
b/Tính GT P khi x\(\frac{2}{2+\sqrt{3}}\)