Đáp án:
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Bài `1`
`e,`
`|x-1/4| =3/7`
`↔` \(\left[ \begin{array}{l}x-\dfrac{1}{4}=\dfrac{3}{7}\\x-\dfrac{1}{4}=\dfrac{-3}{7}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{3}{7}+\dfrac{1}{4}\\x=\dfrac{-3}{7}+\dfrac{1}{4}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{19}{28}\\x=\dfrac{-5}{28}\end{array} \right.\)
Vậy `x=19/28` hoặc `x=(-5)/28`
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`f,`
`|3x + 1/4| = 6/7`
`↔` \(\left[ \begin{array}{l}3x+\dfrac{1}{4}=\dfrac{6}{7}\\3x+\dfrac{1}{4}=\dfrac{-6}{7}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}3x=\dfrac{6}{7}-\dfrac{1}{4}\\3x=\dfrac{-6}{7}-\dfrac{1}{4}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}3x=\dfrac{17}{28}\\3x=\dfrac{-31}{28}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{17}{28}÷3\\x=\dfrac{-31}{28}÷3\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{17}{84}\\x=\dfrac{-31}{84}\end{array} \right.\)
Vậy `x=17/84` hoặc `x=(-31)/84`
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Bài `2`
Có : `hat{A_2} = hat{A_4}` (2 góc đối đỉnh)
mà `hat{A_2} = 40^o`
`-> hat{A_4} =40^o`
Có : `hat{A_2} + hat{A_1} = 180^o` (2 góc kề bù)
`-> hat{A_1} = 180^o - hat{A_2}`
`-> hat{A_1} = 180^o - 40^o`
`-> hat{A_1} = 140^o`
Có : `hat{A_1} = hat{A_3}` (2 góc đối đỉnh)
mà `hat{A_1} = 140^o`
`-> hat{A_3} = 140^o`
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Do $A//B$
`-> hat{A_2} = hat{B_1}` (2 góc đồng vị)
mà `hat{A_2} =40^o`
`-> hat{B_1} =40^o`
Do $A//B$
`-> hat{A_3} = hat{B_2}` (2 góc đồng vị)
mà `hat{A_3} =140^o`
`-> hat{B_2} = 140^o`
Do $A//B$
`-> hat{A_1} = hat{B_4}` (2 góc đồng vị)
mà `hat{A_1} = 140^o`
`-> hat{B_4} = 140^o`
Có : `hat{B_1} = hat{B_3}` (2 góc đối đỉnh)
mà `hat{B_1} = 40^o`
`-> hat{B_3} = 40^o`
Vậy ...
