Tìm x biết : (1/1.3+1/3.5+...+1/19.21).420-(0,4.(7,5-2.5.x))=212
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a, Đề sai hả bạn ??
b, \(\dfrac{\left(1,16-x\right).5,25}{\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right).2\dfrac{2}{17}}=75\%\)
\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{95}{9}-\dfrac{29}{4}\right).\dfrac{36}{17}}=\dfrac{75}{100}\)
\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{380}{36}-\dfrac{261}{36}\right).\dfrac{36}{17}}=\dfrac{3}{4}\)
\(\dfrac{\left(1,16-x\right).5,25}{\dfrac{119}{36}.\dfrac{36}{17}}=\dfrac{3}{4}\)
\(\dfrac{\left(1,16-x\right).5,25}{7}=\dfrac{3}{4}\)
=> \(\left[\left(1,16-x\right).5,25\right].4=3.7\)
\(\left[\left(1,16-x\right).5,25\right].4=21\)
( 1,16 - x ) . 5,25 = 21/4
1,16 - x = 21/4 : 5,25
1,16 - x = 1
x = 1,16 - 1
x = 0,16
Vậy x = 0,16
c, \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(\dfrac{1}{2}.\dfrac{20}{21}.420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(200-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)
\(0,4.\left(7,5-2,5x\right):0,25=200-212\)
\(0,4.\left(7,5-2,5x\right):0,25=-12\)
0,4 . ( 7,5 - 2,5x ) = -12 . 0,25
0,4 . ( 7,5 - 2,5x ) = -3
7,5 - 2,5x = -3 :0,4
7,5 - 2,5x = -7,5
2,5x = 7,5-(-7,5)
2,5x = 15
x = 6
Vậy x = 6
Vậy x = 51
câu a chắc mk nhìn ko rõ vì mk cận mà ko đeo kính, ghi sai đề
đặt A = 1/1.3 + 1/3.5 + ..+ 1/19.21
2A = 2/1.3 + 2/3.5 +..+2/19.21
2A = 3-1/1.3 +...+5-3/3.5 +... + 21-19/19.21
2A = 1/1 - 1/3 + 1/3 -1/5 +...+1/19 -1/21
2A = 20/21
A = 10/21
r thế vào tính là xng
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)
\(\frac{x}{14}=\frac{16}{21}\)
\(\Rightarrow x\cdot=21=14\cdot16\)
\(\Rightarrow x\cdot21=224\)
\(\Rightarrow x=\frac{224}{21}\)
ĐK : 51x \(\ge0\Rightarrow x\ge0\)
Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)
Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)
<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)
<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)
<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)
Vậy x = 50/101
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)