Toán Lớp 8: Tìm x: $4-x = 2(x-4)^2$ $4(x^2)(x-2)+2x=4$

Question

Toán Lớp 8:  Tìm x: $4-x = 2(x-4)^2$ $4(x^2)(x-2)+2x=4$

maianhnhiquynh · Answer

a) 4-x=2(x-4)^2 <=> 2(x-4)^2-(4-x)=0 <=> 2(x-4)^2+(x-4)=0 <=> (x-4)[2(x-4)+1]=0 <=> (x-4)(2x-8+1)=0 <=> [(x-4=0),(2x-7=0):} <=> ( left[ begin{array}{l}x=4 x= dfrac{7}{2} end{array} right. ) Vậy x in {7/2;4} b) 4x^2(x-2)+2x=4 <=> 4x^2(x-2)+2x-4=0 <=> 4x^2(x-2)+2(x-2)=0 <=> (4x^2+2)(x-2)=0 <=> x-2=0 (Vì 4x^2+2 ge 2>0 AAx) <=> x=2 Vậy x=2

dantam45 · Answer

Giải đáp:   a)  ( left[  begin{array}{l} x = 4   x =  dfrac{7}{2}  end{array}  right. )   b) x=2   Lời giải và giải thích chi tiết:    ( begin{array}{l} a)4 - x = 2{ left( {x - 4}  right)^2}     to 2{ left( {x - 4}  right)^2} + x - 4 = 0     to  left( {x - 4}  right) left( {2 left( {x - 4}  right) + 1}  right) = 0     to  left[  begin{array}{l} x - 4 = 0   2x - 8 + 1 = 0  end{array}  right.     to  left[  begin{array}{l} x = 4   x =  dfrac{7}{2}  end{array}  right.   b)4{x^2} left( {x - 2}  right) + 2x - 4 = 0     to 4{x^2} left( {x - 2}  right) + 2 left( {x - 2}  right) = 0     to  left( {x - 2}  right) left( {4{x^2} + 2}  right) = 0     to x - 2 = 0 left( {do:4{x^2} + 2 > 0 forall x}  right)     to x = 2  end{array} )