CMR : \(\overline{a5^2}=\left[a.\left(a+1\right)\right]25\)
Ví dụ như : 15^2 = [ 1 . ( 1 + 1 )]25 = 225
CMR : \(81^{10}-27^{13}-9^{21}⋮225\)
Mình làm như vậy đúng k ạ
\(\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)
\(3^{40}-3^{39}-3^{42}\)
\(3^{39}\left(3-1-27\right)\)
\(3^{39}.\left(-25\right)\)
\(3^{37}.3^2.\left(-25\right)\)
\(3^{37}\left(9.-25\right)\)
\(3^{37}.\left(-225\right)⋮225\)
Đúng ạk
Năm nay đứa nào cũng thấp điểm hết ròi =.= hazz
Cái chỗ : \(3^{37}\left(9.-25\right)\) phải là \(3^{37}.9.\left(-25\right)=3^{37}.\left(-225\right)\) nhé
=.=
CMR: A=B
A=\(\frac{\left(3\frac{2}{15}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)
B=\(\frac{1,2:\left(1\frac{1}{5}.1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
4) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
5) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
6) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp mình với, cần gấp ạ
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
4) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
5) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp vs ạ
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
Tính :
\(A=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right)........\left(1-\frac{1}{196}\right).\left(1-\frac{1}{225}\right)\)
\(\left|A\right|=\left[{}\begin{matrix}A\left(A\ge0\right)\\-A\left(A< 0\right)\end{matrix}\right.\) đúng không nhỉ các bạn . Còn nếu biết a là 1 số rõ ràng ví dụ như -3 , -4 hoặc số chưa rõ như x-3 thì phải xét 2 TH ạ!
\(\left|A\right|=\left\{{}\begin{matrix}A\left(A\ge0\right)\\-A\left(A< 0\right)\end{matrix}\right.\) đúng không nhỉ các bạn . Còn nếu biết a là 1 số rõ ràng ví dụ như -3 , -4 hoặc số chưa rõ như x-3 thì phải xét 2 TH ạ!
Bài 1:
1/\(\left(-\dfrac{25}{13}\right)+\left(-\dfrac{19}{17}\right)+\dfrac{12}{13}+\left(-\dfrac{25}{17}\right)\) 6/ \(2\dfrac{2}{15}.\dfrac{9}{17}.\dfrac{3}{32}:\left(-\dfrac{3}{17}\right)\)
2/\(\dfrac{1}{2}-\left(-\dfrac{1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}\) 7/\(\left(\dfrac{-3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
3/\(\left(-\dfrac{3}{7}\right).\dfrac{5}{11}+\left(-\dfrac{5}{14}\right).\dfrac{5}{11}\) 8/\(\left(-\dfrac{1}{3}\right).\left(-\dfrac{15}{19}\right).\dfrac{38}{45}\)
4/\(\left(-\dfrac{5}{11}\right).\dfrac{7}{15}.\dfrac{11}{-5}.\left(-30\right)\) 9/\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{19.20}\)
5/\(\left(-\dfrac{5}{9}\right).\dfrac{3}{11}+\left(-\dfrac{13}{18}\right).\dfrac{3}{11}\) 10/\(\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-......-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)
Tính giá trị của biểu thức:
a) (-7,05 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\left(-\dfrac{1}{5}\right)\) : (-3)
b) \(\left(\dfrac{3}{25}-1,12\right)\) : \(\dfrac{3}{7}\) : [\(\left(3\dfrac{1}{2}\right)\) - \(\left(3\dfrac{2}{3}\right)\) : \(\dfrac{1}{14}\)]
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)