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13 tháng 1 2016

Nhiều thế bạn tách từng câu ra mik giải cho (olm ko dc trừ điểm câu này của e)

13 tháng 1 2016

Phần b bạn tự làm nhé, chỉ cần quy đồng lên lấy MC = 105 là được mà

Phần a mình giải ntn:

PT \(\Leftrightarrow\) \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)

\(\Leftrightarrow x=-100\)

 

7 tháng 1 2016

a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)

=> \(\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)

=> \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

=> \(\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Ta có : \(\frac{1}{98}+\frac{1}{97}\ne\frac{1}{96}+\frac{1}{95}\) => \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

=> \(x+100=0\)

=> \(x=-100\)

25 tháng 2 2020

giup minh voi cac bạn

12 tháng 3 2020

a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

12 tháng 3 2020

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999

AH
Akai Haruma
Giáo viên
7 tháng 2 2020

a)

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)

$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$

Do đó $x-23=0\Rightarrow x=23$

b)

PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$

$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$

$\Rightarrow x+100=0\Rightarrow x=-100$

AH
Akai Haruma
Giáo viên
7 tháng 2 2020

c)

PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$

Do đó $x+2005=0\Rightarrow x=-2005$

d)

PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)

\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)

\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$

Do đó $300-x=0\Rightarrow x=300$

2 tháng 2 2017

(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94

[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3

[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3

(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94

(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)

(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0

(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0

Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0

=>x+100=0

=>x=-100

k mk nha khong hieu noi mk nha.

2 tháng 2 2017

1/3x-1/2=(3/5-4x)15/7

1/3x-1/2=9/7-60/7x

1/3x+60/7x=1/2+9/7

187/21x=25/14

x=75/374

k mk nha ban.

a) Ta có: 3x(x-1)=(x-1)(x+2)

⇔3x(x-1)-(x-1)(x+2)=0

⇔(x-1)(3x-x-2)=0

⇔(x-1)(2x-1)=0

⇔2(x-1)2=0

mà 2≠0

nên (x-1)2=0

⇔x-1=0

hay x=1

Vậy: x=1

b) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

c) Ta có: \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+2\right)\)

\(\Leftrightarrow\frac{x}{2}+\frac{1}{2}+\frac{x}{4}+\frac{3}{4}=3-\frac{x}{2}-1\)

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{5}{4}-\frac{-x}{2}-2=0\)

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2}-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+x-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+\frac{4x}{4}-\frac{3}{4}=0\)

\(\Leftrightarrow5x-3=0\)

\(\Leftrightarrow5x=3\)

hay \(x=\frac{3}{5}\)

Vậy: \(x=\frac{3}{5}\)

d) Ta có: \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

\(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

nên x+100=0

hay x=-100

Vậy: x=-100

https://i.imgur.com/u6zkAVa.jpg
14 tháng 2 2020

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

\(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!