1 a $\frac{148-x}{25}$ + $\frac{169-x}{23}$ + $\frac{186-x}{21}$ + $\frac{199-x}{19}$ =10 b |4x-$\frac{1}{3}$ |= $\frac{2}{5}$ 2 a tìm n ∈ Z để

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
\dfrac{{148 – x}}{{25}} + \dfrac{{169 – x}}{{23}} + \dfrac{{186 – x}}{{21}} + \dfrac{{199 – x}}{{19}} = 10\\
 \Leftrightarrow \left( {\dfrac{{148 – x}}{{25}} – 1} \right) + \left( {\dfrac{{169 – x}}{{23}} – 2} \right) + \left( {\dfrac{{186 – x}}{{21}} – 3} \right) + \left( {\dfrac{{199 – x}}{{19}} – 4} \right) = 0\\
 \Leftrightarrow \dfrac{{\left( {148 – x} \right) – 25}}{{25}} + \dfrac{{\left( {169 – x} \right) – 2.23}}{{23}} + \dfrac{{\left( {186 – x} \right) – 3.21}}{{21}} + \dfrac{{\left( {199 – x} \right) – 4.19}}{{19}} = 0\\
 \Leftrightarrow \dfrac{{123 – x}}{{25}} + \dfrac{{123 – x}}{{23}} + \dfrac{{123 – x}}{{21}} + \dfrac{{123 – x}}{{19}} = 0\\
 \Leftrightarrow \left( {123 – x} \right)\left( {\dfrac{1}{{25}} + \dfrac{1}{{23}} + \dfrac{1}{{21}} + \dfrac{1}{{19}}} \right) = 0\\
 \Leftrightarrow 123 – x = 0\\
 \Leftrightarrow x = 123\\
b,\\
\left| {4x – \dfrac{1}{3}} \right| = \dfrac{2}{5} \Leftrightarrow \left[ \begin{array}{l}
4x – \dfrac{1}{3} = \dfrac{2}{5}\\
4x – \dfrac{1}{3} =  – \dfrac{2}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{{60}}\\
x =  – \dfrac{1}{{60}}
\end{array} \right.\\
2,\\
a,\\
P = \dfrac{{2n – 3}}{{n + 4}} = \dfrac{{2.\left( {n + 4} \right) – 11}}{{n + 4}} = 2 – \dfrac{{11}}{{n + 4}}\\
P \in Z \Leftrightarrow \dfrac{{11}}{{n + 4}} \in Z \Leftrightarrow n + 4 \in \left\{ { \pm 1; \pm 11} \right\}\\
 \Rightarrow n \in \left\{ { – 15; – 5; – 3;7} \right\}\\
b,\\
a – b = 2\left( {a + b} \right) = a:b\\
a – b = 2\left( {a + b} \right)\\
 \Leftrightarrow 2\left( {a + b} \right) – \left( {a – b} \right) = 0\\
 \Leftrightarrow a + 3b = 0\\
 \Leftrightarrow a =  – 3b\\
a – b = a:b\\
 \Leftrightarrow \left( { – 3b} \right) – b = \left( { – 3b} \right):b\\
 \Leftrightarrow  – 4b =  – 3\\
 \Leftrightarrow b = \dfrac{3}{4} \Rightarrow a =  – 3b =  – \dfrac{9}{4}
\end{array}\)