Tìm x |X+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0 28/08/2021 Bởi Gabriella Tìm x |X+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0
`|x+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0` Vì :`|x + 3/4| ; |y – 1/5| ; |x+y+z|` $≥$ $0$ $∀$ $x;y;z$ `⇒ |x+3/4 |= |y – 1/5 | = |x+y+z| = 0` $⇒$ $\left \{ {{x + \dfrac{3}{4} = 0} \atop {y – \dfrac{1}{5}=0}} \atop {x + y+ z =0} \right.$ $⇒$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-3}{4} + \dfrac{1}{5} + z=0} \right.$ $⇔$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-11}{20} + z=0} \right.$ $⇔$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {z = \dfrac{11}{20}} \right.$ Vậy `(x;y;z)=({-3}/4;1/5;{11}/{20})` Bình luận
Đáp án: Giải thích các bước giải: Ta có `|x+ 3/4 |>=0` `| y – 1/5|>=0` `| x+ y +z | >= 0` `=>|x+ 3/4 |+ | y – 1/5| + | x+ y +z | >= 0` Mà `|x+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0` `=>{x+3/4=0` `{y-1/5=0` `{x+y+z=0` `=>{x=(-3)/4` `{y=1/5` `{z=0-x-y=3/4-1/5=11/20` Bình luận
`|x+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0`
Vì :`|x + 3/4| ; |y – 1/5| ; |x+y+z|` $≥$ $0$ $∀$ $x;y;z$
`⇒ |x+3/4 |= |y – 1/5 | = |x+y+z| = 0`
$⇒$ $\left \{ {{x + \dfrac{3}{4} = 0} \atop {y – \dfrac{1}{5}=0}} \atop {x + y+ z =0} \right.$
$⇒$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-3}{4} + \dfrac{1}{5} + z=0} \right.$
$⇔$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {\dfrac{-11}{20} + z=0} \right.$
$⇔$ $\left \{ {{x= – \dfrac{3}{4} } \atop {y = \dfrac{1}{5}}} \atop {z = \dfrac{11}{20}} \right.$
Vậy `(x;y;z)=({-3}/4;1/5;{11}/{20})`
Đáp án:
Giải thích các bước giải:
Ta có
`|x+ 3/4 |>=0`
`| y – 1/5|>=0`
`| x+ y +z | >= 0`
`=>|x+ 3/4 |+ | y – 1/5| + | x+ y +z | >= 0`
Mà `|x+ 3/4 |+ | y – 1/5| + | x+ y +z | = 0`
`=>{x+3/4=0`
`{y-1/5=0`
`{x+y+z=0`
`=>{x=(-3)/4`
`{y=1/5`
`{z=0-x-y=3/4-1/5=11/20`