Bài 5: Giải các phương trình sau :
a) \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}-6}{5}\)
b) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
c) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
e) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) b) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
c) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi
Câu 6. Giải các phương trình sau:
a, x+\(\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\); b, \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}}{5}-6\)
Câu 7. Giải các phương trình sau:
a, \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\); b, \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4+++==}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c, \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\); d, \(\frac{201-6}{99}+\frac{203-6}{97}=\frac{205-x}{95}+3=0\)
e, \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\); f, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
g, \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\); h, \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
i, \(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\);
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
Bài 1
a,(2x+1)2/5 - (x-1)2/3=7x2-14x-5/15
b,(x+10)(x+4)/12 - (x-4)(2-x)/4=(x+10)(x-2)/3
c,(x-2)2/3 - (2x-3)(2x+3)/8 + (x-4)2/6=0
Bài 2
a,x-23/24 + x-23/25=x-23/26 + x-23/27
b,(x+2/98 + 1)+(x+3/97 + 1)=(x+4/96 + 1)+(x+5/95 + 1)
c,x+1/2004 + x+2/2003=x+3/2002 + x+4/2001
d,201-x/99 + 203-x/97=205-x/95 + 3=0
e,x-45/55 + x-47/53=x-55/45 + x-53/47
f,x+1/9 + x+2/8=x+3/7 + x+4/6
g,x+2/98 +x+4/96=x+6/94 + x+8/92
h,2-x/2002 - 1=1-x/2003 - x/2004
i,x2-10x-29/1971 + x2-10x -27/1973=x2-10x-1971/29 + x2-10x-1973/27
Hơi khó hiểu thông cảm ạ
Baif 1 : giải phương trình
a. \(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
b. \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)\left(\dfrac{x+5}{95}+1\right)\)
c. \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
d. \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
e.\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
f. \(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
h. \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
g. \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
h.
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)
Suy ra: 2004 - x = 0
Vậy x = 2004
\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )
\(\Leftrightarrow x=23\)
Vậy pt có tập nghiệm S = { 23 }
\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { 100 }
\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\)
\(\Leftrightarrow x=300\)
Vậy pt có tập nghiệm S = { 300 }
\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Vậy pt có tập nghiệm S = { 10 }
\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
\(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
Vậy pt có tập nghiệm S = { 2004 }
\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { -100 }
a.
\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
Vì: \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\)
Suy ra x - 23 = 0
\(\Leftrightarrow x=23\)
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm
a) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
b)\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
giải phương trình(tiếp)
\(\frac{5}{3x+2}=2x-1\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3\)
\(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x-\frac{x-3}{2}}{5}-x+1\)
\(\frac{\frac{1}{2}+a}{a-\frac{1}{2}}-\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a\left(3a+1\right)}{a^2-\left(\frac{1}{2}\right)^2}\)
\(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm