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\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)
Vậy phương trình có nghiệm \(x=2004\)
\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)
a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-105\right)=0;\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\ne0\)
\(\Leftrightarrow x=105\)
b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)=0\)
\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)
\(\Leftrightarrow x=50\)
Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)
\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)
\(\Leftrightarrow x=1999\)(thỏa mãn)
Vậy \(x=1999\)
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-27}{1973}-1\right)+\left(\dfrac{x^2-10x-29}{1971}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1973}+\dfrac{x^2-10x-2000}{1971}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)=0\) vì \(\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)\ne0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy: Giá trị x thỏa mãn là: \(x=-40;50\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
vì \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
Nên \(x^2-10x-2000=0\)
<=> \(x^2-50x+40x-2000=0\)
<=> \(x\left(x-50\right)+40\left(x-50\right)=0\)
<=> \(\left(x-50\right)\left(x+40\right)=0\)
<=> \(x=50\) hoặc \(x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)
\(\dfrac{x-1}{13}-\dfrac{2x-13}{15}=\dfrac{3x-15}{27}-\dfrac{2x-27}{29}\)
\(\Leftrightarrow\dfrac{x-1}{13}-1-\dfrac{2x-13}{15}-1=\dfrac{3x-15}{27}-1-\dfrac{2x-27}{29}-1\)
\(\Leftrightarrow\dfrac{x-1-13}{13}-\dfrac{2x-13-15}{15}=\dfrac{3x-15-27}{27}-\dfrac{4x-27-29}{29}\)
\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2x-24}{15}=\dfrac{3x-42}{27}-\dfrac{4x-56}{29}\)
\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2\left(x-14\right)}{15}-\dfrac{3\left(x-14\right)}{27}-\dfrac{4\left(x-14\right)}{29}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\right)=0\)
\(\Leftrightarrow x-14=0\) ( Vì: \(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\ne0\))
\(\Leftrightarrow x=14\)
b: \(\Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)+\left(\dfrac{21-x}{29}+1\right)=0\)
=>50-x=0
hay x=50
c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)
=>x-2003=0
hay x=2003
Ta có: \(A=\frac{108}{27}\cdot\frac{146}{29}-\frac{54}{27}\cdot\frac{202}{29}-\frac{16}{29}\)
\(=4\cdot\frac{146}{29}-2\cdot\frac{202}{29}-\frac{16}{29}\)
\(=\frac{584}{29}-\frac{404}{29}-\frac{16}{29}\)
\(=\frac{164}{29}\)
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
giúp mình với :(((((