Toán Lớp 8: Tìm X bt;
d/ (2x-3)^2-(x+5)^2=0
giúp!!!!!!!!!!!!!
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Toán Lớp 8: Tìm X bt;
d/ (2x-3)^2-(x+5)^2=0
giúp!!!!!!!!!!!!!
Comments ( 2 )
Giải đáp:
Vậy x = 8 ; x = {-2}/3
Lời giải và giải thích chi tiết:
d)
(2x – 3)^2 – (x + 5)^2 = 0
[(2x – 3) – (x + 5)][(2x – 3) + (x + 5)] = 0
(2x – 3 – x – 5)(2x – 3 + x + 5) = 0
(x – 8)(3x + 2) = 0
=> \(\left[ \begin{array}{l}x-8=0\\3x+2=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=8\\x = \dfrac{- 2}{3}\end{array} \right.\)
Vậy x = 8 ; x = {-2}/3
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* Dùng HĐT 3: a^2 – b^2 = (a – b)(a + b)
#SunHee
d, (2x-3)^2 – (x+5)^2 = 0
[( 2x-3) + (x+5)][( 2x-3) – (x+5)] =0
( 2x -3 + x+5)(2x-3-x-5)=0
( 3x +2)(x-8)=0
=>\(\left[ \begin{array}{l}3x+2=0\\x-8=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}3x=-2\\x=8\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=\frac{-2}{3}\\x=8\end{array} \right.\)
Vậy x ∈ {-2/3 ; 8}
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– Dùng HĐT số 3: a^2 – b^2 = (a+b)(a-b)