\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}\frac{x+7}{29}\)
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TM
29 tháng 12 2016
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
<=>\(\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)
<=>\(\frac{x+1+35}{35}+\frac{x+3+33}{33}=\frac{x+5+31}{31}+\frac{x+7+29}{29}\)
<=>\(\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
<=>\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
<=>\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\Rightarrow x+36=0\Rightarrow x=-36\)
TK
5 tháng 4 2016
a/ (X+1)/35+1+(x+3)/33+1 =(x+5)/31+(x+7)/29+1+1
=>(x+36)/35+(x+36)/33-(x+36)/31-(x+36)/27=0
=>(X+36)(1/35+1/33-1/31-1/29)=0
=> x+36=0(vì c=vế 2 luôn luôn khác 0)
=>x=-36
b/ CMTT câu a
trừ tung phân số cho 1 ta được x=2004
9 tháng 8 2016
Cộng 3 vào cả 2 vế và chuyển vế xong đặt nhân tử x+50 ra ngoài ta được :
(x+50)(1/39+1/37+1/35-1/33-1/31-1/29)=0
vì (1/39+1/37+1/35-1/33-1/31-1/29) khác 0
=> x+50=0
=> x=-50
Nhớ k mình nhé
Chúc bạn học tốt
8 tháng 1 2020
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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
29 tháng 3 2020
b, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1994}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1994}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)=0\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có tập nghiệm là \(S=\left\{2004\right\}\)
13 tháng 3 2020
a) Sửa đề: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
Ta có: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)
nên x+36=0
hay x=-36
Vậy: x=-36
LT
7
13 tháng 8 2017
\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)
\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)
\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)
\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)
VÌ \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)
\(\Rightarrow x+60=0\)
\(\Rightarrow x=-60\)
NC
1
5 tháng 1 2020
Bài làm
\(\frac{x+19}{27}-\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
\(\Leftrightarrow\left(\frac{x+19}{27}+1\right)-\left(\frac{x+17}{29}+1\right)=\left(\frac{x+15}{31}+1\right)-\left(\frac{x+13}{33}+1\right)\)
\(\Leftrightarrow\frac{x+46}{27}-\frac{x+46}{29}=\frac{x+46}{31}-\frac{x+46}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}=\left(x+46\right).\frac{1}{31}-\left(x+46\right).\frac{1}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}-\left(x+46\right).\frac{1}{31}+\left(x+46\right).\frac{1}{33}=0\)
\(\Leftrightarrow\left(x+46\right)\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)=0\)
Mà \(\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)>0\forall x\)
\(\Leftrightarrow x+46=0\)
\(\Leftrightarrow x=-46\)
Vậy phương trình trên có tập nghiệm S = { -46 }
# Học tốt #
17 tháng 7 2016
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
NV
17 tháng 7 2016
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
đề chính xác là như này đúng k cậu=)đề k rõ ràng k aii giúp đc nhé!
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\Leftrightarrow\left(\frac{x+1}{35}+1\right)+\left(\frac{x+3}{33}+1\right)=\left(\frac{x+5}{31}+1\right)+\left(\frac{x+7}{29}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+35}{35}\right)+\left(\frac{x+3+33}{33}\right)=\left(\frac{x+5+31}{31}\right)+\left(\frac{x+7+29}{29}\right)\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\Leftrightarrow\left(x+36\right).\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0.\)
\(\Leftrightarrow x+36=0\)
\(\Leftrightarrow x=0-36\)
\(\Leftrightarrow x=-36.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-36\right\}.\)
Chúc bạn học tốt!