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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Giải cho mik câu 58 sgk toán 8 trang 39 vs. Mik đang cần gấp mai mik phải noopk bài r. Ai làm nhanh cho 5 sao, lm đúng cho trlhn

Tháng Sáu 10, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Giải cho mik câu 58 sgk toán 8 trang 39 vs. Mik đang cần gấp mai mik phải noopk bài r. Ai làm nhanh cho 5 sao, lm đúng cho trlhn

Comments ( 1 )

  1. huenthanh97
    huenthanh97
    10 Tháng Sáu, 2022 at 2:44 sáng
    Reply
    a)
    $\eqalign{  &  = \left[ {{9 \over {x\left( {x + 3} \right)\left( {x – 3} \right)}} + {1 \over {x + 3}}} \right]:\left[ {{{x – 3} \over {x\left( {x + 3} \right)}} – {x \over {3\left( {x + 3} \right)}}} \right]  \cr  &  = {{9 + x\left( {x – 3} \right)} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}}:{{3\left( {x – 3} \right) – {x^2}} \over {3x\left( {x + 3} \right)}} = {{{x^2} – 3x + 9} \over {x\left( {x + 3} \right)\left( {x – 3} \right)}}.{{3x\left( {x + 3} \right)} \over {3x – 9 – {x^2}}}  \cr  &  = {{3\left( {{x^2} – 3x + 9} \right)} \over {\left( {3 – x} \right)\left( {{x^2} – 3x + 9} \right)}} = {3 \over {3 – x}} \cr}$
    b)
    $= {{2x + 4 – 2x + 4} \over {\left( {x – 2} \right)\left( {x + 2} \right)}}.{{{{\left( {x + 2} \right)}^2}} \over 8} = {8 \over {\left( {x – 2} \right)\left( {x + 2} \right)}}.{{{{\left( {x + 2} \right)}^2}} \over 8} = {{x + 2} \over {x – 2}}$
    c)
    $\eqalign{  &  = {{9{x^2} + 3x + 2x – 6{x^2}} \over {\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}.{{{{\left( {1 – 3x} \right)}^2}} \over {2x\left( {3x + 5} \right)}} = {{x\left( {3x + 5} \right)} \over {\left( {1 – 3x} \right)\left( {1 + 3x} \right)}}.{{{{\left( {1 – 3x} \right)}^2}} \over {2x\left( {3x + 5} \right)}}  \cr  &  = {{1 – 3x} \over {2\left( {1 + 3x} \right)}} \cr}$
    d)
    $\left( {{x \over {{x^2} – 25}} – {{x – 5} \over {{x^2} + 5x}}} \right):{{2x – 5} \over {{x^2} + 5x}} + {x \over {5 – x}}$
    $\eqalign{ &  = \left[ {{x \over {\left( {x + 5} \right)\left( {x – 5} \right)}} – {{x – 5} \over {x\left( {x + 5} \right)}}} \right]:{{2x – 5} \over {x\left( {x + 5} \right)}} + {x \over {5 – x}}  \cr  &  = {{{x^2} – {{\left( {x – 5} \right)}^2}} \over {x\left( {x + 5} \right)\left( {x – 5} \right)}}.{{x\left( {x + 5} \right)} \over {2x – 5}} + {x \over {5 – x}}  \cr  &  = {{{x^2} – {x^2} + 10x – 25} \over {\left( {x – 5} \right)\left( {2x – 5} \right)}} + {x \over {5 – x}} = {{5\left( {2x – 5} \right)} \over {\left( {x – 5} \right)\left( {2x – 5} \right)}} – {x \over {x – 5}}  \cr  &  = {5 \over {x – 5}} – {x \over {x – 5}} = {{5 – x} \over {x – 5}} = {{ – \left( {x – 5} \right)} \over {x – 5}} =  – 1 \cr}$
    e)
    $\left( {{{{x^2} + xy} \over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y \over {{x^2} + {y^2}}}} \right):\left( {{1 \over {x – y}} – {{2xy} \over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} \right)$
    $\eqalign{  &  = \left[ {{{{x^2} + xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}} + {y \over {{x^2} + {y^2}}}} \right]:\left[ {{1 \over {x – y}} – {{2xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)}}} \right]  \cr  &  = {{{x^2} + xy + y\left( {x + y} \right)} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}:{{{x^2} + {y^2} – 2xy} \over {\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)}}  \cr  &  = {{{x^2} + xy + xy + {y^2}} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}.{{\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)} \over {{{\left( {x – y} \right)}^2}}}  \cr  &  = {{{{\left( {x + y} \right)}^2}} \over {\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}.{{\left( {{x^2} + {y^2}} \right)\left( {x – y} \right)} \over {{{\left( {x – y} \right)}^2}}} = {{x + y} \over {x – y}} \cr}$

    toan-lop-8-giai-cho-mik-cau-58-sgk-toan-8-trang-39-vs-mik-dang-can-gap-mai-mik-phai-noopk-bai-r

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222-9+11+12:2*14+14 = ? ( )

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