Tìm x, y, z biết z: a, | 4 – 2x| + | x – 2| = 3 -x b, x – 1 / 2 = y -3 / 4 = z -2 / 3 và x – 3y + 4z = 4 c, 3|x| + 5 / 3 = 3|y| – 1 / 5 = 3 – z / 7 v

Giải thích các bước giải:

\(\begin{array}{l}
a)2\left| {2 – x} \right| + \left| {x – 2} \right| = 3 – x\\
2\left| {x – 2} \right| + \left| {x – 2} \right| = 3 – x\\
3\left| {x – 2} \right| = 3 – x\\
TH1:x \ge 2\\
3\left( {x – 2} \right) = 3 – x\\
4x = 9\\
x = \dfrac{9}{4}\left( {tm} \right)\\
TH2:x < 2\\
3\left( {2 – x} \right) = 3 – x\\
2x = 3\\
x = \dfrac{3}{2}\left( {tm} \right)\\
b)\dfrac{{x – 1}}{2} = \dfrac{{y – 3}}{4} = \dfrac{{z – 2}}{3}\\
 = \dfrac{{x – 1 – 3\left( {y – 3} \right) + 4\left( {z – 2} \right)}}{{2 – 3.4 + 4.3}} = \dfrac{{x – 3y + 4z}}{2} = \dfrac{4}{2} = 2\\
 \Rightarrow x – 1 = 4 \Rightarrow x = 5\\
y – 3 = 8 \Rightarrow y = 11\\
z – 2 = 6 \Rightarrow z = 8\\
c)2\left| x \right| + 7\left| y \right| + 3z =  – 14\\
 \Rightarrow 6\left| x \right| + 21\left| y \right| + 9z =  – 42\\
\dfrac{{3\left| x \right| + 5}}{3} = \dfrac{{3\left| y \right| – 1}}{5} = \dfrac{{3 – z}}{7}\\
 = \dfrac{{2\left( {3\left| x \right| + 5} \right) + 7\left( {3\left| y \right| – 1} \right) – 9\left( {3 – z} \right)}}{{2.3 + 5.7 – 9.7}}\\
 = \dfrac{{6\left| x \right| + 21\left| y \right| + 9z – 24}}{{ – 22}} = \dfrac{{ – 42 – 24}}{{ – 22}} = 3\\
 \Rightarrow 3\left| x \right| + 5 = 9 \Leftrightarrow 3\left| x \right| = 4 \Leftrightarrow x =  \pm \dfrac{4}{3}\\
3\left| y \right| – 1 = 15 \Rightarrow y =  \pm \dfrac{{16}}{3}\\
3 – z = 21 \Leftrightarrow z =  – 18
\end{array}\)