tìm x a) 29-x /21+ 27-x/23 + 23-x/27 + 21-x /27 =0 b) x – 5/100 + x-4/101 + x-3/102 = x-100/5 + x-101/4 +x-102/3

Đáp án:

b. x=105

Giải thích các bước giải:

\(\begin{array}{l}
a.\dfrac{{29 – x}}{{21}} + \dfrac{{27 – x}}{{23}} + \dfrac{{23 – x}}{{25}} + \dfrac{{21 – x}}{{27}} = 0\\
 \to \dfrac{{29 – x}}{{21}} + 1 + \dfrac{{27 – x}}{{23}} + 1 + \dfrac{{23 – x}}{{25}} + 1 + \dfrac{{21 – x}}{{27}} + 1 = 4\\
 \to \dfrac{{50 – x}}{{21}} + \dfrac{{50 – x}}{{23}} + \dfrac{{50 – x}}{{25}} + \dfrac{{50 – x}}{{27}} = 0\\
 \to \left( {50 – x} \right)\left( {\dfrac{1}{{21}} + \dfrac{1}{{23}} + \dfrac{1}{{25}} + \dfrac{1}{{27}}} \right) = 0\\
 \to 50 – x = 0\\
 \to x = 50\\
b.\dfrac{{x – 5}}{{100}} + \dfrac{{x – 4}}{{101}} + \dfrac{{x – 3}}{{102}} = \dfrac{{x – 100}}{5} + \dfrac{{x – 101}}{4} + \dfrac{{x – 102}}{3}\\
 \to \dfrac{{x – 5}}{{100}} – 1 + \dfrac{{x – 4}}{{101}} – 1 + \dfrac{{x – 3}}{{102}} – 1 = \dfrac{{x – 100}}{5} – 1 + \dfrac{{x – 101}}{4} – 1 + \dfrac{{x – 102}}{3} – 1\\
 \to \dfrac{{x – 105}}{{100}} + \dfrac{{x – 105}}{{101}} + \dfrac{{x – 105}}{{102}} = \dfrac{{x – 105}}{5} + \dfrac{{x – 105}}{4} + \dfrac{{x – 105}}{3}\\
 \to \dfrac{{x – 105}}{{100}} + \dfrac{{x – 105}}{{101}} + \dfrac{{x – 105}}{{102}} – \dfrac{{x – 105}}{5} – \dfrac{{x – 105}}{4} – \dfrac{{x – 105}}{3} = 0\\
 \to \left( {x – 105} \right)\left( {\dfrac{1}{{100}} + \dfrac{1}{{101}} + \dfrac{1}{{102}} – \dfrac{1}{5} – \dfrac{1}{4} – \dfrac{1}{3}} \right) = 0\\
 \to x – 105 = 0\\
 \to x = 105
\end{array}\)