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

Toán Lớp 8: khai triển hằng đẳng thức để tính

Toán Lớp 8: khai triển hằng đẳng thức để tính A, (3x – 2/3)^3
b, 16x^2 – 81x^2
c, 1/25 – 36y^2

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    $a)\; \left(3x-\dfrac23\right)^2=(3x)^2-2.3x.\dfrac23+\left(\dfrac23\right)^2=9x^2-4x+\dfrac{4}{9}$
    $b)\; 16x^2-81x^2=(4x)^2-(9x)^2=(4x-9x)(4x+9x)=-5x.13x=-65x^2$
    $c)\; \dfrac1{25}-36y^2=\left(\dfrac15\right)^2-(6y)^2=\left(\dfrac{1}{5}-6y\right)\left(\dfrac{1}{5}+6y\right)$

  2. ~rai~
    \(a)\left(3x-\dfrac{2}{3}\right)^3\\=(3x)^3-3.(3x)^2.\dfrac{2}{3}+3.3x.\left(\dfrac{2}{3}\right)^2-\left(\dfrac{2}{3}\right)^3\\=27x^3-18x^2+4x-\dfrac{8}{27}.\\b)16x^2-81x^2\\=-65x^2.\\\text{Sửa đề:}16x^2-81y^2\\=(4x)^2-(9y)^2\\=(4x-9y)(4x+9y).\\c)\dfrac{1}{25}-36y^2\\=\left(\dfrac{1}{5}\right)^2-(6y)^2\\=\left(\dfrac{1}{5}-6y\right)\left(\dfrac{1}{5}+6y\right).\)

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