\(\dfrac{x+3}{1995}\)+\(\dfrac{x+1}{1997}\)=-2
Giải nhanh giùm mk nha!!!
\(\dfrac{x+1}{1998}\)+ \(\dfrac{x+2}{1997}\)=\(\dfrac{x+3}{1996}\)+\(\dfrac{x+4}{1995}\)
=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)
=>x+1999=0
=>x=-1999
giải các phương trình sau
\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x +3}{1996}+\dfrac{x+4}{1995}\)
Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T
\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)
\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)
\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(=\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
⇔\(x+1999=0\)
Vậy \(x=-1999\)
giải các phương Trình sau
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
Giải các phương trình:
a) \(\dfrac{x+2001}{5}+\dfrac{x+1999}{7}+\dfrac{x+1997}{9}+\dfrac{x+1995}{11}=-4;\)
b) \(\dfrac{x-15}{100}+\dfrac{x-10}{105}+\dfrac{x-100}{110}=\dfrac{x-100}{15}+\dfrac{x-105}{10}+\dfrac{x-110}{5}.\)
a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)
=>x+2006=0
=>x=-2006
b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)
=>x-105=0
=>x=105
Tính giá trị biểu thức sau bằng cách thuận tiện nhất :
a, 3/5 + 6/11 + 7/13 + 2/5 + 16/11 + 19/13
b, 1995/1997 x 1990/1993 x 1997/1994 x 1993/1995 x 997/995
Ai nhanh mình tk nhưng phải tk cho mk nha !
\(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
\(=1+2+2\)
\(=5\)
a
[3/5+2/5]+[7/13+19/13]+[6/11+16/11]=5
Tính nhanh:
\(\dfrac{1988\times1996+1997+1985}{1997\times1996-1995\times1996}\)
\(\dfrac{1988\:×\:1996\:+\:1997\:+\:1985}{1997\:×\:1996\:-\:1995\:×\:1996}\) (rút bỏ các phần tử, mẫu giống nhau)
= \(\dfrac{1988\:+\:1985}{1995\:×\:1996}\)
= (còn lại tự tính)
RÚT GỌN BIỂU THỨC SAU
\(\left(x+\dfrac{1}{3}x+\dfrac{1}{9}\right)\left(x-\dfrac{1}{3}\right)-\left(x-\dfrac{1}{3}\right)^2\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
Giari ptr
a/2x(3x-1)=6x^2-13
b/\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)
Giups mk vs ạ ai nhanh mk tick nha ><
a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)
b: =>2x-2x-1=x-6x
=>-5x=-1
hay x=1/5
Lời giải:
a.
$2x(3x-1)=6x^2-13$
$\Leftrightarrow 6x^2-2x=6x^2-13$
$\Leftrightarrow 2x=13$
$\Leftrightarrow x=\frac{13}{2}$
b.
$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$
$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$
$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$
Giải phương trình
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)
Giúp mk nhanh nha mk đag cần gấp nhé
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(< =>\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)=\left(\dfrac{x+5}{61}+1\right)+\left(\dfrac{x+7}{59}+1\right)\)
\(< =>\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)
\(< =>\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(< =>x+66=0< =>x=-66\)
Vậy tập nghiệm của phương trình đã cho là: S={-66}
\(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)
\(< =>\dfrac{x+60}{31}-\dfrac{x+60}{33}=\dfrac{x+60}{43}-\dfrac{x+60}{45}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\)
Mà: \(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\)
\(=>x+60=0< =>x=-60\)
Vậy tập nghiệm của phương trình đã cho là: s={-60}