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Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a) \(\left(2-\frac{3}{2}\right)\left(2-\frac{4}{3}\right)\left(2-\frac{5}{4}\right)\left(2-\frac{6}{4}\right)\)
\(=\frac{1}{3}\left(-\frac{4}{3}+2\right)\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\left(-\frac{6}{4}+2\right)\)
\(=\frac{1.2.3\left(2-\frac{3}{2}\right)}{2.3.4}\)
\(=\frac{1.3\left(2-\frac{3}{2}\right)}{3.4}\)
\(=\frac{1.\left(2-\frac{3}{2}\right)}{4}\)
\(=\frac{2-\frac{3}{4}}{4}\)
\(=\frac{1}{2.4}\)
\(=\frac{1}{8}\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
\(=\frac{8028024\left(\frac{2003}{2004}+\frac{2004}{2003}\right)}{8028025}\)
\(=\frac{8028024.\frac{8028025}{4014012}}{8028025}\)
\(=\frac{16056050}{8028025}\)
= 2
a. \(25^3:5^2\)
\(=\left(5^2\right)^3:5^2\)
\(=5^6:5^2=5^4\)
b. \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21-\left(2+6\right)}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
\(a,25^3:5^2\)
=\(\left(5^2\right)^3:5^2\)
=\(5^6:5^2\)
=\(5^4\)
\(b,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
=\(\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(c,3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
=\(3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(d,\left(-\frac{7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}\)
\(=\left(-\frac{7}{4}\cdot\frac{8}{5}\right)\cdot\frac{11}{16}\)
\(=-\frac{14}{5}\cdot\frac{11}{16}\)
\(=-\frac{77}{40}\)
\(e,\frac{2}{3}+\frac{1}{3}\cdot\frac{-6}{10}\)
\(=\frac{2}{3}-\frac{1}{5}\)
\(=\frac{7}{15}\)
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
Bài 1:
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+1986}\right)\)
Nhận xét: \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Do đó: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+1986}\right)\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{1985\cdot1988}{1986\cdot1987}=\frac{1\cdot4\cdot1988}{1986\cdot3}=\frac{3976}{2979}\)
Bài 2:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
\(\Rightarrow\frac{4\cdot4^5}{3\cdot3^5}\cdot\frac{6\cdot6^5}{2\cdot2^5}=2^x\)\(\Rightarrow\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=2^x\)
\(\Rightarrow\frac{\left(2^2\right)^6}{3^6}\cdot\frac{\left(2\cdot3\right)^6}{2^6}=2^x\)\(\Rightarrow\frac{2^{12}}{3^6}\cdot\frac{2^6\cdot3^6}{2^6}=2^x\)
\(\Rightarrow\frac{2^6\cdot3^6\cdot2^{12}}{2^6\cdot3^6}=2^x\)\(\Rightarrow2^{12}=2^x\Rightarrow x=12\)
2m - 2n = 256 = 28 \(\Rightarrow\)2n . ( 2m-n - 1 ) = 28
dễ thấy m \(\ne\)n , ta xét 2 trường hợp :
a) nếu m - n = 1 thì từ ( 1 ) ta có : 2n . ( 2 - 1 ) = 28 . suy ra : n = 8, m = 9
b) nếu m - n \(\ge\)2 thì 2m-n - 1 là 1 số lẻ lớn hơn 1 nên vế trái của ( 1 ) chứa thừa số nguyên tố lẻ khi phân tích ra thừa số nguyên tố. còn vế phải của ( 1 ) chỉ chứa thừa số nguyên tố 2. Mâu thuẫn
Vậy n = 8 , m = 9 là đáp số bài trên
đặt A = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
3A = \(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
3A - A = 2A = \(1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)
biểu thức trong dấu ngoặc nhỏ hơn \(\frac{1}{2}\)( tự chứng minh ) nên 2A < 1 + \(\frac{1}{2}\)
\(\Rightarrow A< \frac{3}{4}\)