(x-100):24+(x-98):26+(x-98):28=3
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21 tháng 12 2020
a) Ta có: \(\left(4x-1\right)^2=\left(1-4x\right)^2\)
\(\Leftrightarrow\left(4x-1\right)^2-\left(1-4x\right)^2=0\)
\(\Leftrightarrow\left(4x-1-1+4x\right)\left(4x-1+1-4x\right)=0\)
\(\Leftrightarrow0\cdot x=0\)(luôn đúng)
Vậy: \(x\in R\)
b) Ta có: \(\dfrac{x-100}{24}+\dfrac{x-98}{26}+\dfrac{x-96}{28}=3\)
\(\Leftrightarrow\dfrac{x-100}{24}-1+\dfrac{x-98}{26}-1+\dfrac{x-96}{28}-1=0\)
\(\Leftrightarrow\dfrac{x-124}{24}+\dfrac{x-124}{26}+\dfrac{x-124}{28}=0\)
\(\Leftrightarrow\left(x-124\right)\cdot\left(\dfrac{1}{24}+\dfrac{1}{26}+\dfrac{1}{28}\right)=0\)
mà \(\dfrac{1}{24}+\dfrac{1}{16}+\dfrac{1}{28}>0\)
nên x-124=0
hay x=124
Vậy: x=124
NE
4
OD
22 tháng 6 2017
Ta có:\(\frac{x-100}{24}+\frac{x-98}{26}+\frac{x-96}{28}=3\)
\(\Rightarrow\)\(\frac{91x-100.91}{91.24}+\frac{84x-84.98}{26.84}+\frac{78x-96.78}{78.28}\)
=\(\frac{91x-9100+84x-8232+78x-7488}{2184}\)
=\(\frac{91x+84x+78x-9100-8232-7488}{2184}\)
=\(\frac{x\left(91+84+78\right)-\left(9100+8232+7488\right)}{2184}\)
=\(\frac{x253-24820}{2184}=3\)
\(\Rightarrow\)x253- 24820 =6552
\(\Rightarrow\)x253= 31372
\(\Rightarrow\)x = 124
LM
4 tháng 12 2017
\(\frac{x-100}{24}+\frac{x-98}{+26}+\frac{x-96}{28}=3\)
\(=\frac{\left(x-100\right)}{24}+\frac{\left(x-98\right)}{26}+\frac{\left(x-96\right)}{28}-1=0\)
\(\Leftrightarrow\frac{\left(x-100\right)}{24-1}+\frac{\left(x-98\right)}{26-1}+\frac{\left(x-96\right)}{28-1}=0\)
\(\Leftrightarrow\left(x-124\right)\left(\frac{1}{24}+\frac{1}{26}+\frac{1}{28}\right)=0\)
Vì: \(\frac{1}{24}+\frac{1}{26}+\frac{1}{28}\ne0\)
\(\Rightarrow x-124=0\)
\(\Rightarrow x=124-0\)
\(\Rightarrow x=124\)