Toán Lớp 10: 1 giải các pt sau a, ∛(5-3x) = -2 b, √(-x ²+3x+4) = x-5 c, √(3x+4) + √(x-3) = √(6x+1)

Question

Toán Lớp 10:  1 giải các pt sau a,  ∛(5-3x) = -2 b, √(-x ²+3x+4) = x-5 c,  √(3x+4) + √(x-3) =  √(6x+1)

hoquyly · Answer

Giải đáp+Lời giải và giải thích chi tiết: a) root{3}{5-3x}=-2 <=>5-3x=-8 <=>3x=13 <=>x=13/3 Vậy S={13/3} ____________________________________________________ b) sqrt{(-x^2+3x+4)}=x-5 <=>-x^2+3x+4=x^2-10x+25 (x>=5) <=>2x^2-13x+21=0 <=>(x-3)(2x-7)=0 <=> ( left[ begin{array}{l}x-3=0 2x-7=0 end{array} right. ) <=> ( left[ begin{array}{l}x=3 x= dfrac{7}{2} end{array} right. ) Vậy S={3;7/2} ________________________________________________________ c) sqrt{3x+4}+ sqrt{x-3}= sqrt{6x+1} (x>=3) <=>3x+4+x-3+2 sqrt{(3x+4)(x-3)}=6x+1 <=>2 sqrt{(3x+4)(x-3)}=2x <=> sqrt{(3x+4)(x-3)}=x <=>3x^2-9x+4x-12=x^2 <=>2x^2-5x-12=0 <=>(x-4)(2x+3)=0 <=> ( left[ begin{array}{l}x-4=0 2x+3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=4 x= dfrac{-3}{2} end{array} right. ) Vậy S={4;-3/2}