tìm x, biết
a, 25x^2 -160=9
b, (x+4) ^2 – (x-1) (x+1)=16
c, (2x-1)^2 + (x+3)^2 – 5(x-7)(x+7)=0
d, (x-2)^2 – 4 (x^2 – 4) +4 (x+2)^2=0
giúp mk vs . Mk đg gấp . Thanks trước
tìm x, biết a, 25x^2 -160=9 b, (x+4) ^2 – (x-1) (x+1)=16 c, (2x-1)^2 + (x+3)^2 – 5(x-7)(x+7)=0 d, (x-2)^2 – 4 (x^2 – 4) +4 (x+2)^2=0
By Josephine
`a, 25x^2 – 160 = 9`
`⇔ 25x^2 = 169`
`⇔ x^2 = 169/25`
`⇔ x = ± 13/5`
`b, (x+4)^2 – (x-1)(x+1) = 16`
`⇔ x^2 + 8x + 16 – x^2 + 1 = 16`
`⇔ 8x = -1`
`⇔ x = -1/8`
`c, (2x – 1)^2 + (x+3)^2 – 5(x-7)(x+7) = 0`
`⇔ 4x^2 – 4x + 1 + x^2 + 6x + 9 – 5(x^2 – 49) = 0`
`⇔ 5x^2 + 2x + 10 – 5x^2 + 245 = 0`
`⇔ 2x = -255`
`⇔ x = -255/2`
`d, (x-2)^2 – 4(x^2 – 4) + 4(x + 2)^2 = 0`
`⇔ x^2 – 4x + 4 – 4x^2 + 16 + 4(x^2 + 4x + 4) = 0`
`⇔ -3x^2 – 4x + 20 + 4x^2 + 16x + 16 = 0`
`⇔ x^2 + 12x + 36 = 0`
`⇔ (x + 6)^2 = 0`
`⇔ x + 6 = 0`
`⇔ x = -6`
Đáp án:
a/ `x = – \frac{1}{8}`
b/ `x = – \frac{1}{8}`
c/ `x =-\frac{255}{2}`
d/ `x = -6`
Giải thích các bước giải:
a/ `25x^2 – 160 =9`
`<=> 25x^2 – 169 = 0`
`<=> ( 5x – 13 )( 5x + 13 ) = 0`
`<=>` \(\left[ \begin{array}{l}5x – 13 = 0\\5x + 13 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{13}{5}\\x=-\dfrac{13}{5}\end{array} \right.\)
b/ `( x + 4 )^2 – ( x -1)( x + 1 ) = 16`
`<=> x^2 + 8x + 16 – x^2 + 1 = 16`
`<=> 8x = -1`
`<=> x = – \frac{1}{8}`
c/ `( 2x – 1 )^2 + ( x + 3 )^2 – 5( x – 7 )( x + 7 ) = 0`
`<=> 4x^2 – 4x + 1 + x^2 + 6x + 9 – 5( x^2 – 49 ) = 0`
`<=> 4x^2 – 4x + 1 + x^2 + 6x + 9 – 5x^2 + 245 =0`
`<=> 2x = -255`
`<=> x =-\frac{255}{2}`
d/ `( x- 2)^2 – 4( x^2 – 4 ) + 4( x + 2 )^2 = 0`
` <=> [ ( x – 2 ) – 2( x + 2 ) ]^2 = 0`
`<=> ( x – 2 – 2x – 4 )^2 = 0`
`<=> ( -x – 6 )^2= 0`
`<=> x = -6`