Toán Lớp 8: Câu1:(x+3)³-x(3x+1)²+(2x+1)(4x²+1-2x)-3x²=54 Câu 2: 14x²(2x-1)+21x(1-2x)=0 Câu 3 : 5x(x-1)⁵=10x³(1-x)³

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Toán Lớp 8:  Câu1:(x+3)³-x(3x+1)²+(2x+1)(4x²+1-2x)-3x²=54 Câu 2: 14x²(2x-1)+21x(1-2x)=0 Câu 3 : 5x(x-1)⁵=10x³(1-x)³

thucquyenlan · Answer

Giải đáp + Lời giải và giải thích chi tiết: Câu 1 : (x+3)^3-x(3x+1)^2+(2x+1)(4x^2+1-2x)-3x^2=54 <=>x^3+9x^2+27x+27-x(9x^2+6x+1)+(2x+1)(4x^2-2x+1)-3x^2=54 <=>x^3+9x^2+27x+27-9x^3-6x^2-x+(2x)^3+1-3x^2=54 <=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54 <=>(x^3-9x^3+8x^3)+(9x^2-6x^2-3x^2)+(27x-x)+(27+1)=54 <=>26x+28=54 <=>26x=26<=>x=1 Vậy S={1} Câu 2 : 14x^2(2x-1)+21x(1-2x)=0 <=>14x^2(2x-1)-21x(2x-1)=0 <=>(2x-1)(14x^2-21x)=0 <=>(2x-1)7x(2x-3)=0 <=> ( left[ begin{array}{l}2x-1=0 7x=0 2x-3=0 end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{1}{2} x=0 x= dfrac{3}{2} end{array} right. ) Vậy S={1/2;0;3/2} Câu 3 : 5x(x-1)^5=10x^3(1-x)^3 <=>5x(x-1)^5-10x^3(1-x)^3=0 <=>5x(x-1)^5+10x^3(x-1)^3=0 <=>(x-1)^3[5x(x-1)^2+10x^3]=0 <=>(x-1)^3[5x(x^2-2x+1)+10x^3]=0 <=>(x-1)^3(5x^3-10x^2+5x+10x^3)=0 <=>(x-1)^3(15x^3-10x^2+5x)=0 <=>(x-1)^3*5x(3x^2-2x+1)=0 <=>[((x-1)^3=0),(5x=0),(3x^2-2x+1=0):} <=>[(x-1=0),(x=0),(3x^2-2x+1=0):} <=>[(x=1),(x=0),(x=emptyset):} Vậy S={1;0}.