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<=> (x-1)/99-1 + (x-2)/49-2 + (x-7)/31-3 +(x-8)/23-4=0
<=> (x-100)/99 + (x-100)/49 + (x-100)/31 + (x-100)/23=0
<=> (x-100)(1/99 + 1/49 + 1/31 + 1/23)=0
<=> x-100=0(vì 1/99 + 1/49 + 1/31 +1/23)
<=> x=100
Vậy PT có TN S={100}
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
\(\frac{x-12}{21}+\frac{x-10}{23}=\frac{x-8}{25}+\frac{x-6}{27}\)
\(\Leftrightarrow\frac{x-12-21}{21}+\frac{x-10-23}{23}-\frac{x-8-25}{25}-\frac{x-6-27}{27}=0\)
\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}-\frac{x-33}{25}-\frac{x-33}{27}=0\)
\(\Leftrightarrow\left(x-33\right)\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)=0\)
Vif \(\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)\ne0\)
\(\Rightarrow x-33=0\)
\(\Rightarrow x=33\)
\(\frac{x-12}{21}+\frac{x-10}{23}=\frac{x-8}{25}+\frac{x-6}{27}\)
\(\Leftrightarrow\frac{x-12}{21}+1+\frac{x-10}{23}+1=\frac{x-8}{25}+1+\frac{x-6}{27}+1\)
\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}=\frac{x-33}{25}+\frac{x-33}{27}\)
\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}-\frac{x-33}{25}-\frac{x-33}{27}=0\)
\(\Leftrightarrow\left(x-33\right)\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)=0\)
Mà \(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\ne0\)
\(\Rightarrow x-33=0\)
\(\Leftrightarrow x=33\)
Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)
Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)
b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)
=> k2 \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)
=> k = -2;2
+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{z}{7}=2\Rightarrow z=14\)
+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)
\(\frac{z}{7}=2\Rightarrow z=-14\)
Vậy................................
Câu A
X + (X+1) + (X+3) +...+ (X+2003) = 2004
Số số hạng trong tổng 1 + 3 + ... + 2003 là
(2003 - 1) : 2 + 1 = 1002
Tổng dãy 1 + 3 + ... + 2003 là:
(1 + 2003) * 1002 : 2 = 1004004
=> (1003.X) + 1004004 = 2004
=> (1003.X)= 2004 - 1004004
=> 1003.X = - 1002000
X = - 1002000/1003
E chỉ giải đc đến đây thui!!!!!!!!!!!!!!! :)))
x + ( x + 1) + (x + 3) ... + (x + 2003) = 2004
x + x + x + ... + x (có 1003 x) + 1 + 3 + 5 + ... + 2003 = 2004
x . 1003 + 1004004 = 2004
x . 1003 = 2004 - 1004004
x . 1003 = -1002000
x = -1002000 : 1003
x = -999,00299 = ~-999
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-7}{31}-3+\frac{x-8}{23}-4=0\)
\(\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{31}+\frac{x-100}{23}=0\)
\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{31}+\frac{1}{23}\right)=0\)
x-100=0 ( vi 1/99+1/49+1/31+1/23 khác 0)
x=100