4.\(\left(\frac{1}{4}\right)^2+25.\text{[}\text{ }\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\text{]}:\left(\frac{3}{2}\right)^3\)
\(A=\frac{3}{\left(1\text{*}2\right)\text{*}\left(1\text{*}2\right)}+\frac{5}{\left(2\text{*}3\right)\text{*}\left(3\text{*}2\right)}+\frac{7}{\left(3\text{*}4\right)\text{*}\left(3\text{*}4\right)}+...............+\frac{19}{\left(9\text{*}10\right)\text{*}\left(10\text{*}9\right)}\)
theo bài ra ta có
n = 8a +7=31b +28
=> (n-7)/8 = a
b= (n-28)/31
a - 4b = (-n +679)/248 = (-n +183)/248 + 2
vì a ,4b nguyên nên a-4b nguyên => (-n +183)/248 nguyên
=> -n + 183 = 248d => n = 183 - 248d (vì n >0 => d<=0 và d nguyên )
=> n = 183 - 248d (với d là số nguyên <=0)
vì n có 3 chữ số lớn nhất => n<=999 => d>= -3 => d = -3
=> n = 927
\(\text{Tìm }x,\text{ }\text{biết:}\)
\(1\text{)}\text{ }2.\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{2}\right)=\frac{1}{2}x\)
\(2\text{) }-3\left(x-\frac{1}{4}\right)-\frac{1}{3}\left(x+\frac{1}{2}\right)=x\)
\(\text{3) }\frac{3}{2}\left(x-\frac{5}{3}\right)-\frac{4}{5}=x+1\)
\(\text{4) }\frac{1}{6}\left(2.x-3\right)=\frac{1}{2}\left(-x+\frac{1}{4}-\frac{2}{3}\right)\)
\(\text{5) }-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2.x-1\right)\)
( Câu siêu khó dành cho học sinh đội tuyển Toán )
( Giúp với )
Bài 10 :
b) \(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}\)
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
Giải :
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
~~Học tốt~~
\(\text{Giải :}\)
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
\(\text{~~Học tốt~~}\)
Tính nhanh
a) \(\frac{6}{7}:\left(\frac{1}{2}\text{X}\frac{3}{4}\right)-\frac{5}{8}\)
b) 34-2:\(\left(\frac{3}{5}-\frac{1}{2}\right)\)
c) \(\left(4\frac{1}{2}+\frac{1}{2}:5\frac{1}{2}\right)\text{X}\left(3\frac{5}{6}+2\frac{1}{6}\text{x}6\right)\)
\(\left(-2\right)^3.\left(\frac{1}{2}\right)^2+\left(1,75-25\%\right):\frac{3}{\left(-2\right)^4}\)
= -8 . \(\frac{1}{4}\)+ 1,5 : \(\frac{3}{16}\)
= -2+ 8
= 6
Yêu cầu, gợi ý các bạn khác chọn (k) đúng cho mình
( Bài hơi dài , mong mọi người giúp ạ ! )
\(\text{Tính P :}\)
\(P=\frac{0,8:\left(\frac{4}{5}\times1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\times2\frac{2}{17}}+\left(1,2\times0,5\right):\frac{4}{5}\)
tính B=\(\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).....\left(29^4+\frac{1}{4}\right)}{\left(2^{\text{4}}+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)......\left(30^4+\frac{1}{4}\right)}\)
Tính \(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+\text{4}\right)+...+\frac{1}{2016}\left(1+2+...+2016\right)\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+2016\right)\)
\(A=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)
\(A=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)
\(A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)
\(A=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)
Tính:
1. \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2\text{x}}{x^2-1}\)
2. \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2\text{x}}{x^2+x+1}-\frac{6}{x-1}\)
3. \(\frac{5}{2\text{x}^2+6\text{x}}-\frac{4-3\text{x}^2}{x^2-9}-3\)
4. \(\frac{7}{8x^2-18}+\frac{1}{2\text{x}^2+3\text{x}}-\frac{1}{4\text{x}-6}\)
5. \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)