\(\text { Giải hệ phương trình }\left\{\begin{array}{l} 2|x+y|+|x-y|=7 \\ -|x+y|+4|x-y|=10 \text { . } \end{array}\right.\)

A. \(\left(\frac{5}{2} ;-\frac{1}{2}\right) ;\left(-\frac{1}{2} ; \frac{5}{2}\right) ;\left(\frac{1}{2} ;-\frac{5}{2}\right) ;\left(-\frac{5}{2} ; \frac{1}{2}\right) .\)

B. \(\left(\frac{5}{2} ;-\frac{1}{2}\right) ;\left(-\frac{1}{2} ; \frac{5}{2}\right) ;\left(\frac{1}{2} ;\frac{5}{2}\right) ;\left(-\frac{5}{2} ; \frac{1}{2}\right) .\)

C. \(\left(-\frac{1}{2} ; \frac{5}{2}\right) ;\left(\frac{1}{2} ;-\frac{5}{2}\right) ;\left(-\frac{5}{2} ; \frac{1}{2}\right) .\)

D. \(\left(\frac{5}{2} ;-\frac{1}{2}\right) ;\left(-\frac{1}{2} ; \frac{5}{2}\right) ;\left(\frac{1}{2} ;-\frac{5}{2}\right) \)

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Môn: Toán Lớp 10

Chủ đề: Hàm số, đồ thị và ứng dụng

Bài: Phương trình quy về phương trình bậc hai

Lời giải:

Báo sai

\(\text { Đặt }\left\{\begin{array}{l} u=|x+y| \\ v=|x-y| \end{array}, u, v \geq 0\right.\)

Hệ phương trình theo u;v:
\(\left\{\begin{array} { l } { 2 u + v = 7 } \\ { - u + 4 v = 1 0 } \end{array} \Leftrightarrow \left\{\begin{array} { l } { v = 7 - 2 u } \\ { - u + 4 ( 7 - 2 u ) = 1 0 } \end{array} \Leftrightarrow \left\{\begin{array}{l} v=3 \\ u=2 \end{array}\right.\right.\right.\)

\(\text { Suy ra }\left\{\begin{array}{l} |x+y|=2 \\ |x-y|=3 \end{array}\right.\)\(\Leftrightarrow\left\{\begin{array}{c} {\left[\begin{array}{l} x+y=2 \\ x+y=-2 \end{array}\right.} \\ |x-y|=3 \end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{l} \left\{\begin{array}{l} y=2-x \\ |x-y|=3 \end{array}\right. \\ \left\{\begin{array}{l} y=-2-x \\ |x-y|=3 \end{array}\right. \end{array}\right.\)

\(\text { Xét hệ }\left\{\begin{array} { l } { y = 2 - x } \\ { | x - y | = 3 } \end{array} \Leftrightarrow \left\{\begin{array} { l } { y = 2 - x } \\ { | x - ( 2 - x ) | = 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} y=2-x \\ {\left[\begin{array}{l} 2 x-2=3 \\ 2 x-2=-3 \end{array}\right.} \end{array}\right.\right.\right.\)

\(\Leftrightarrow\left\{\begin{array}{c} y=2-x \\ {\left[\begin{array}{l} x=\frac{5}{2} \\ x=-\frac{1}{2} \end{array}\right.} \end{array}\right.\)\(\left[ \begin{array}{l} \left\{ {\begin{array}{*{20}{l}} {x = \frac{5}{2}}\\ {y = - \frac{1}{2}} \end{array}} \right.\\ \left\{ {\begin{array}{*{20}{l}} {x = - \frac{1}{2}}\\ {y = \frac{5}{2}} \end{array}} \right. \end{array} \right.\)

\(\text { Xét hệ }\left\{\begin{array} { l } { y = - 2 - x } \\ { | x - y | = 3 } \end{array} \Leftrightarrow \left\{\begin{array} { l } { y = - 2 - x } \\ { | x - ( - 2 - x ) | = 3 } \end{array} \Leftrightarrow \left\{\begin{array}{l} y=-2-x \\ {\left[\begin{array}{l} 2 x+2=3 \\ 2 x+2=-3 \end{array}\right.} \end{array}\right.\right.\right.\)\(\Leftrightarrow\left\{\begin{array}{l} y=-2-x \\ {\left[\begin{array}{l} x=\frac{1}{2} \\ x=-\frac{5}{2} \end{array}\right.} \end{array}\right.\)\(\left[ \begin{array}{l} \left\{ {\begin{array}{*{20}{l}} {x = \frac{1}{2}}\\ {y = - \frac{5}{2}} \end{array}} \right.\\ \left\{ {\begin{array}{*{20}{l}} {x = - \frac{5}{2}}\\ {y = \frac{1}{2}} \end{array}} \right. \end{array} \right.\)

\(\text { Vậy hệ phương trình có } 4 \text { nghiệm là }\left(\frac{5}{2} ;-\frac{1}{2}\right) ;\left(-\frac{1}{2} ; \frac{5}{2}\right) ;\left(\frac{1}{2} ;-\frac{5}{2}\right) ;\left(-\frac{5}{2} ; \frac{1}{2}\right) \text { . }\)

\(\text { Giải hệ phương trình }\left\{\begin{array}{l} 2|x+y|+|x-y|=7 \\ -|x+y|+4|x-y|=10 \text { . } \end{array}\right.\)

Lời giải:

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