mong mọi người giúp mình
a)c-23/24+x-23/25=x-23/26=x-23/27
b)x+2/98+x-3/97=x+4/96+x+5/95
c)x-45/55+x-47/53=x-55/49+x-53/47
mong mọi người giúp mình
a)c-23/24+x-23/25=x-23/26=x-23/27
b)x+2/98+x-3/97=x+4/96+x+5/95
c)x-45/55+x-47/53=x-55/49+x-53/47
Đáp án:
c) x=100
Giải thích các bước giải:
a)x−2324+x−2325=x−2326+x−2327→(x−23)(124+125−126−127)=0→x−23=0→x=23b)x+298+x+397=x+496+x+595→x+298+1+x+397+1=x+496+1+x+595+1→x+10098+x+10097=x+10096+x+10095→(x+100)(198+197−196−195)=0→x=−100c)x−4555+x−4753=x−5149+x−5347→x−4555−1+x−4753−1=x−5149−1+x−5347−1→x−10055+x−10053−x−10049−x−10047=0→(x−100)(155+153−149−147)=0→x−100=0→x=100
Đáp án:
c) x=100
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{x – 23}}{{24}} + \dfrac{{x – 23}}{{25}} = \dfrac{{x – 23}}{{26}} + \dfrac{{x – 23}}{{27}}\\
\to \left( {x – 23} \right)\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}} – \dfrac{1}{{26}} – \dfrac{1}{{27}}} \right) = 0\\
\to x – 23 = 0\\
\to x = 23\\
b)\dfrac{{x + 2}}{{98}} + \dfrac{{x + 3}}{{97}} = \dfrac{{x + 4}}{{96}} + \dfrac{{x + 5}}{{95}}\\
\to \dfrac{{x + 2}}{{98}} + 1 + \dfrac{{x + 3}}{{97}} + 1 = \dfrac{{x + 4}}{{96}} + 1 + \dfrac{{x + 5}}{{95}} + 1\\
\to \dfrac{{x + 100}}{{98}} + \dfrac{{x + 100}}{{97}} = \dfrac{{x + 100}}{{96}} + \dfrac{{x + 100}}{{95}}\\
\to \left( {x + 100} \right)\left( {\dfrac{1}{{98}} + \dfrac{1}{{97}} – \dfrac{1}{{96}} – \dfrac{1}{{95}}} \right) = 0\\
\to x = – 100\\
c)\dfrac{{x – 45}}{{55}} + \dfrac{{x – 47}}{{53}} = \dfrac{{x – 51}}{{49}} + \dfrac{{x – 53}}{{47}}\\
\to \dfrac{{x – 45}}{{55}} – 1 + \dfrac{{x – 47}}{{53}} – 1 = \dfrac{{x – 51}}{{49}} – 1 + \dfrac{{x – 53}}{{47}} – 1\\
\to \dfrac{{x – 100}}{{55}} + \dfrac{{x – 100}}{{53}} – \dfrac{{x – 100}}{{49}} – \dfrac{{x – 100}}{{47}} = 0\\
\to \left( {x – 100} \right)\left( {\dfrac{1}{{55}} + \dfrac{1}{{53}} – \dfrac{1}{{49}} – \dfrac{1}{{47}}} \right) = 0\\
\to x – 100 = 0\\
\to x = 100
\end{array}\)