1,X + 2/3 = 4/5 2,X -2/7 = 7/21 3, -X -3/4 = -8/11 4,11/12 – (2/5 +X ) = 2/3 5,1/2 – (1/3 +3/4)bé hơn hoặc bằng x bé hơn hoặc bằng 1/24 -(1/8 -1/3)

1,X + 2/3 = 4/5
2,X -2/7 = 7/21
3, -X -3/4 = -8/11
4,11/12 – (2/5 +X ) = 2/3
5,1/2 – (1/3 +3/4)bé hơn hoặc bằng x bé hơn hoặc bằng 1/24 -(1/8 -1/3)

0 bình luận về “1,X + 2/3 = 4/5 2,X -2/7 = 7/21 3, -X -3/4 = -8/11 4,11/12 – (2/5 +X ) = 2/3 5,1/2 – (1/3 +3/4)bé hơn hoặc bằng x bé hơn hoặc bằng 1/24 -(1/8 -1/3)”

  1. Giải thích các bước giải:

    1) $x+\dfrac{2}{3}=\dfrac{4}{5}\\ \Rightarrow x = \dfrac{4}{5}-\dfrac{2}{3}\\ \Rightarrow x = \dfrac{12}{15}-\dfrac{10}{15}\\\Rightarrow x=\dfrac{2}{15}$

    2) $x- \dfrac{2}{7}=\dfrac{7}{21}\\⇒x=\dfrac{7}{21}+\dfrac{2}{7}\\⇒x=\dfrac{7}{21}+\dfrac{6}{21}\\⇒x=\dfrac{13}{21}$

    3) $-x-\dfrac{3}{4}=\dfrac{-8}{11}\\⇒ \dfrac{-3}{4}-\dfrac{-8}{11}=x\\⇒\dfrac{-33}{44}-\dfrac{-32}{44}=x\\⇒\dfrac{-1}{44}=x$

    4) $\dfrac{11}{12}-\bigg(\dfrac{2}{5}+x\bigg)=\dfrac{2}{3}\\⇒ \dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\⇒x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\\⇒x= \dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}\\⇒x=\dfrac{-3}{20}$

    5) $\dfrac{1}{2}-\bigg(\dfrac{1}{3}+\dfrac{3}{4}\bigg) \le x \le \dfrac{1}{24}-\bigg(\dfrac{1}{8}-\dfrac{1}{3}\bigg)\\⇒ \dfrac{6}{12}-\bigg(\dfrac{4}{12}+\dfrac{9}{12}\bigg) \le x \le \dfrac{1}{24}-\bigg(\dfrac{3}{24}-\dfrac{8}{24}\bigg)\\⇒\dfrac{-7}{12} \le x\le \dfrac{3}{12}$

    `=>x∈{(-7)/12;(-1)/2;(-5)/12;(-1)/3;(-1)/4;(-1)/6;(-1)/12;0;1/12;1/6;3/12}`

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  2. Đáp án:

     

    Giải thích các bước giải:

     $1) x= \dfrac{4}{5}- \dfrac{2}{3}$

    $x= \dfrac{2}{15}$

    $2) x= \dfrac{7}{21}+ \dfrac{2}{7}$

    $x= \dfrac{7}{21}+ \dfrac{6}{21}$

    $x= \dfrac{13}{21}$

    $3) -x= \dfrac{-8}{11}+ \dfrac{3}{4}$

    $-x= \dfrac{1}{44}$

    $⇒ x= -\dfrac{1}{44}$

    $4) \dfrac{11}{12}- \dfrac{2}{5}- x= \dfrac{2}{3}$

    $-x= \dfrac{2}{3}- \dfrac{11}{12}+ \dfrac{2}{5}$

    $-x= \dfrac{3}{20}$

    $⇒ x= -\dfrac{3}{20}$

    $5) \dfrac{1}{2}- \dfrac{13}{12}≤ x≤ \dfrac{1}{24}- \dfrac{-5}{24}$

    $\dfrac{-7}{12}≤ x≤ \dfrac{1}{24}$

    $⇒ \dfrac{-14}{24}≤ x≤ \dfrac{1}{24}$

    $⇒ x∈ \dfrac{-14}{24}; \dfrac{-13}{24}; \dfrac{-12}{24};….; \dfrac{1}{24}$

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