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TM
5
S
13 tháng 9 2017
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>x = 12
D
8 tháng 7 2018
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\)
ML
1
XO
27 tháng 1 2020
Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+2}{98}\)
\(\Rightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+2}{98}+1\)
\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right).\left(\frac{1}{94}+\frac{1}{92}\right)\)
=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}\right)-\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)=0\)
=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\Rightarrow x+100=0\left(\text{vì }\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
=> x = - 100
Vậy x = - 100
NT
3
KY
15 tháng 9 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{92}>\frac{1}{94}>\frac{1}{96}>\frac{1}{98}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
15 tháng 9 2019
\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0.Ma:\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}< 0\Rightarrow x=-100\)
VM
3 tháng 2 2020
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=0-100\)
\(\Leftrightarrow x=-100.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=0+89\)
\(\Leftrightarrow x=89.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)
Chúc bạn học tốt!
31 tháng 7 2022
1: \(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+6}{94}+1\right)+\left(\dfrac{x+8}{92}+1\right)\)
=>x+100=0
hay x=-100
2: \(\Leftrightarrow x\cdot\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>3/4x+5/4=-1/3x+7/3
=>13/12x=13/12
hay x=1
T
3
29 tháng 12 2019
Câu này tớ giải hơn 10 lần rồi cậu ( ko xàm :)
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
Do đó \(x+100=0\Leftrightarrow x=-100\)
Vậy pt có nghiệm : x=-100
XO
29 tháng 12 2019
Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
=> \(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
=> \(\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
=> x + 100 = 0
=> x = - 100
Vậy x = - 100
C
2
BA
31 tháng 10 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\frac{100+x}{98}+\frac{100+x}{96}-\frac{100+x}{94}-\frac{100+x}{92}=0\)
\(\Rightarrow\left(100+x\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\ne0\)
\(\Rightarrow100+x=0\)
\(\Rightarrow x=-100\)
C
31 tháng 10 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}< \frac{1}{96}< \frac{1}{94}< \frac{1}{92}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)
Vậy \(x+100=0\Leftrightarrow x=-100\)
\(\frac{x-96}{2}+\frac{x-88}{4}+\frac{x-76}{6}+\frac{x-60}{8}=14\)
\(\frac{x-96}{2}-2+\frac{x-88}{4}-3+\frac{x-76}{6}-4+\frac{x-60}{8}-5=0\)
\(\frac{x-96}{2}-\frac{4}{2}+\frac{x-88}{4}-\frac{12}{4}+\frac{x-76}{6}-\frac{24}{6}+\frac{x-60}{8}-\frac{40}{8}=0\)
\(\frac{x-96-4}{2}+\frac{x-88-12}{4}+\frac{x-76-24}{6}+\frac{x-60-40}{8}=0\)
\(\frac{x-100}{2}+\frac{x-100}{4}+\frac{x-100}{6}+\frac{x-100}{8}=0\)
\((x-100)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-100=0\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}>0\right)\)
\(\Rightarrow x=100\)