Toán Lớp 7: ||x+1/3|-8|=5/3 mn nên nhớ có 2 dấu giá trị tuyệt đối ở ngoài

Question

Toán Lớp 7:  ||x+1/3|-8|=5/3 mn nên nhớ có 2 dấu giá trị tuyệt đối ở ngoài

buithaianh98 · Answer

$  $   | |x+1/3| - 8| = 5/3   Trường hợp 1 :   -> |x+1/3| -8=5/3   ->|x+1/3|=5/3 + 8   -> |x+1/3|=29/3   ->  ( left[  begin{array}{l}x+ dfrac{1}{3}= dfrac{29}{3}  x+ dfrac{1}{3}= dfrac{-29}{3} end{array}  right. ) $  $ ->  ( left[  begin{array}{l}x= dfrac{29}{3}- dfrac{1}{3}  x= dfrac{-29}{3}- dfrac{1}{3} end{array}  right. ) $  $ ->  ( left[  begin{array}{l}x= dfrac{28}{3}  x=-10 end{array}  right. )    Trường hợp 2 :   -> |x+1/3|-8=(-5)/3   -> |x+1/3|=(-5)/3 + 8   -> |x+1/3|=19/3   ->  ( left[  begin{array}{l}x+ dfrac{1}{3}= dfrac{19}{3}  x+ dfrac{1}{3}= dfrac{-19}{3} end{array}  right. ) $  $ ->  ( left[  begin{array}{l}x= dfrac{19}{3}- dfrac{1}{3}  x= dfrac{-19}{3}- dfrac{1}{3} end{array}  right. ) $  $ ->  ( left[  begin{array}{l}x=6  x= dfrac{-20}{3} end{array}  right. )    Vậy x &isin; {28/3; -10; 6;(-20)/3}

diemchau · Answer

Giải đáp: x in {-10;-20/3;6;28/3}. Lời giải và giải thích chi tiết: ||x+1/3|-8|=5/3 TH1: |x+1/3|-8=5/3 <=>|x+1/3|=8+5/3 <=>|x+1/3|=29/3 (⇔ left[ begin{array}{l}x+ dfrac13= dfrac{29}{3} x+ dfrac13=- dfrac{29}{3} end{array} right. ) (⇔ left[ begin{array}{l}x= dfrac{28}{3} x=- dfrac{30}{3}=-10 end{array} right. ) TH2' |x+1/3|-8=-5/3 <=>|x+1/3|=19/3 (⇔ left[ begin{array}{l}x+ dfrac13= dfrac{19}{3} x+ dfrac13=- dfrac{19}{3} end{array} right. ) (⇔ left[ begin{array}{l}x= dfrac{18}{3}=6 x=- dfrac{20}{3} end{array} right. ) Vậy x in {-10;-20/3;6;28/3}.