Toán Lớp 8: bài 4: tìm x biết
a)4x(x-5)-(x-1)(4x-3)=10
b) (2x-1)(x-2)-(x+3)(2x-7)=3
c) (x+3)(x-4)+(x-1)(x+1)=10
d)8x(x-3)-8(x-1)(x+1)=20
giúp mik nhé !
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Toán Lớp 8: bài 4: tìm x biết
a)4x(x-5)-(x-1)(4x-3)=10
b) (2x-1)(x-2)-(x+3)(2x-7)=3
c) (x+3)(x-4)+(x-1)(x+1)=10
d)8x(x-3)-8(x-1)(x+1)=20
giúp mik nhé !
Comments ( 2 )
⇔x²-4x+3x-12+x²-1-10=0
⇔(x²+x²)+(-4x+3x)-(12+1+10)=0
⇔2x²-x-23=0
⇔2(x²-1/2x-23/2)=0
⇔x²-1/2x-23/2=0
⇔x²-1/2x+1/16-185/16=0
⇔x²-1/2x+1/16=185/16
⇔x²-2.x. 1/4+(1/4)^2=185/16
⇔(x-1/4)^2=($\dfrac{\sqrt{185}}{4}$)^2
⇔$\left[\begin{matrix} x-\dfrac{1}{4}=\dfrac{\sqrt{185}}{4}\\ x-\dfrac{1}{4}=-\dfrac{\sqrt{185}}{4}\end{matrix}\right.$
⇔$\left[\begin{matrix} x=\dfrac{\sqrt{185}}{4}+\dfrac{1}{4}\\ x=-\dfrac{\sqrt{185}}{4}+\dfrac{1}{4}\end{matrix}\right.$
⇔$\left[\begin{matrix} x=\dfrac{\sqrt{185}+1}{4}\\ x=\dfrac{1-\sqrt{185}}{4}\end{matrix}\right.$
Vậy x\in{$\dfrac{\sqrt{185}+1}{4}$;$\dfrac{1-\sqrt{185}}{4}$}
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