Tìm x biết a) |x+2, 8|=3, 5 b) |x+1/4|-1/3=5/6 08/08/2021 Bởi Maria Tìm x biết a) |x+2, 8|=3, 5 b) |x+1/4|-1/3=5/6
Đáp án: a) |x+2, 8|=3, 5 TH1 :|x+2, 8|=3, 5–>x+2,8=3,5–>x=3,5-2,8–>x=0,7 TH2 : |x+2, 8|=-3, 5–>x+2,8=-3,5–>x=-3,5-2,8=-6,3 b) |x+1/4|-1/3=5/6 |x+1/4|=5/6+1/3 |x+1/4|=7/6 TH1 : |x+1/4|=7/6–>x=7/6-1/4–>x=11/12 TH2 : |x+1/4|=-7/6–>x=-7/6-1/4–>x=-17/12 chúc em hc tốt mong em cho chị ctlhn :33 Bình luận
Đáp án: Giải thích các bước giải: $a) |x+2,8|=3,5$ $⇔\left[ \begin{array}{1}x+2,8=3,5\\x+2,8=-3,5\end{array} \right.$ $⇔\left[ \begin{array}{1}x=0,7\\x=-6,3\end{array} \right.$ Vậy $x∈\{0,7 ; -6,3\}$ $b,\left |x+\dfrac{1}{4} \right |-\dfrac{1}{3}=\dfrac{5}{6}$ $⇔\left |x+\dfrac{1}{4} \right |=\dfrac{5}{6}+\dfrac{1}{3}$ $⇔\left |x+\dfrac{1}{4} \right |=\dfrac{7}{6}$ $⇔\left[ \begin{array}{1}x+\dfrac{1}{4}=\dfrac{7}{6}\\x+\dfrac{1}{4}=-\dfrac{7}{6}\end{array} \right.$ $⇔\left[ \begin{array}{1}x=\dfrac{11}{12}\\x=-\dfrac{17}{12}\end{array} \right.$ Vậy $x∈\left \{\dfrac{11}{12} ; -\dfrac{17}{12} \right \}$ Bình luận
Đáp án:
a) |x+2, 8|=3, 5
TH1 :|x+2, 8|=3, 5–>x+2,8=3,5–>x=3,5-2,8–>x=0,7
TH2 : |x+2, 8|=-3, 5–>x+2,8=-3,5–>x=-3,5-2,8=-6,3
b) |x+1/4|-1/3=5/6
|x+1/4|=5/6+1/3
|x+1/4|=7/6
TH1 : |x+1/4|=7/6–>x=7/6-1/4–>x=11/12
TH2 : |x+1/4|=-7/6–>x=-7/6-1/4–>x=-17/12
chúc em hc tốt
mong em cho chị ctlhn :33
Đáp án:
Giải thích các bước giải:
$a) |x+2,8|=3,5$
$⇔\left[ \begin{array}{1}x+2,8=3,5\\x+2,8=-3,5\end{array} \right.$
$⇔\left[ \begin{array}{1}x=0,7\\x=-6,3\end{array} \right.$
Vậy $x∈\{0,7 ; -6,3\}$
$b,\left |x+\dfrac{1}{4} \right |-\dfrac{1}{3}=\dfrac{5}{6}$
$⇔\left |x+\dfrac{1}{4} \right |=\dfrac{5}{6}+\dfrac{1}{3}$
$⇔\left |x+\dfrac{1}{4} \right |=\dfrac{7}{6}$
$⇔\left[ \begin{array}{1}x+\dfrac{1}{4}=\dfrac{7}{6}\\x+\dfrac{1}{4}=-\dfrac{7}{6}\end{array} \right.$
$⇔\left[ \begin{array}{1}x=\dfrac{11}{12}\\x=-\dfrac{17}{12}\end{array} \right.$
Vậy $x∈\left \{\dfrac{11}{12} ; -\dfrac{17}{12} \right \}$