Toán Lớp 8: Giải các pt sau 4) (x+6)(5-x) = -2(5-x)(7x+8) 5) (x-2)(7-3x) = (3x-7)(8x+32) 6) (2-x)(x+1)= (x-2)(3x+5) TLHN+5SAO+CÁM ƠN

Question

Toán Lớp 8:  Giải các pt sau 4) (x+6)(5-x) = -2(5-x)(7x+8) 5) (x-2)(7-3x) = (3x-7)(8x+32)            6) (2-x)(x+1)= (x-2)(3x+5) TLHN+5SAO+CÁM ƠN

maitrucloan · Answer

Giải đáp:   4)S={5;-22/15}   5)S={7/3;-10/3}     6)S={2;-3/2}     Lời giải và giải thích chi tiết:   4)(x+6)(5-x)=-2(5-x)(7x+8)   &hArr;(x+6)(5-x)+2(5-x)(7x+8)=0   &hArr;(5-x)[x+6+2(7x+8)]=0   &hArr;(5-x)(x+6+14x+6)=0   &hArr;(5-x)(15x+22)=0   &hArr; ( left[  begin{array}{l}5-x=0  15x+22=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=5  15x=-22 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=5  x= dfrac{-22}{15} end{array}  right. )    Vậy S={5;-22/15}     5)(x-2)(7-3x)=(3x-7)(8x+32)     &hArr;(x-2)(7-3x)-(3x-7)(8x+32)=0     &hArr;(x-2)(7-3x)+(7-3x)(8x+32)=0     &hArr;(7-3x)(x-2+8x+32)=0     &hArr;(7-3x)(9x+30)=0    &hArr; ( left[  begin{array}{l}7-3x=0  9x+30=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}3x=7  9x=-30 end{array}  right. )   &hArr; ( left[  begin{array}{l}x= dfrac{7}{3}  x=- dfrac{10}{3} end{array}  right. )     Vậy S={7/3;-10/3}       6)(2-x)(x+1)=(x-2)(3x+5)       &hArr;(2-x)(x+1)-(x-2)(3x+5)=0       &hArr;(2-x)(x+1)+(2-x)(3x+5)=0       &hArr;(2-x)(x+1+3x+5)=0       &hArr;(2-x)(4x+6)=0     &hArr; ( left[  begin{array}{l}2-x=0  4x+6=0 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=2  4x=-6 end{array}  right. )   &hArr; ( left[  begin{array}{l}x=2  x=- dfrac{3}{2} end{array}  right. )     Vậy S={2;-3/2}

ngoclanhoa · Answer

*** Lời giải chi tiết *** 4) (x+6)(5-x)=-2(5-x)(7x+8) <=>(x+6)(5-x)+2(5-x)(7x+8)=0 <=>(5-x)[x+6+2(7x+8)]=0 <=>(5-x)(x+6+14x+16)=0 <=>(5-x)(15x+22)=0 <=> ( left[ begin{array}{l}5-x=0 15x+22=0 end{array} right. ) <=> ( left[ begin{array}{l}x=5 x=- dfrac{22}{15} end{array} right. ) Vậy tập nghiệm phương trình là : S={5;-(22)/(15)} 5) (x-2)(7-3x)=(3x-7)(8x+32) <=>(x-2)(7-3x)-(3x-7)(8x+32)=0 <=>(x-2)(7-3x)+(7-3x)(8x+32)=0 <=>(7-3x)(x-2+8x+32)=0 <=>(7-3x)(9x+30)=0 <=> ( left[ begin{array}{l}7-3x=0 9x+30=0 end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{7}{3} x=- dfrac{10}{3} end{array} right. ) Vậy phương trình có tập nghiệm là : S={(7)/(3);-(10)/(3)} 6) (2-x)(x+1)=(x-2)(3x+5) <=>(2-x)(x+1)-(x-2)(3x+5)=0 <=>(2-x)(x+1)+(2-x)(3x+5)=0 <=>(2-x)(x+1+3x+5)=0 <=>(2-x)(4x+6)=0 <=> ( left[ begin{array}{l}2-x=0 4x+6=0 end{array} right. ) <=> ( left[ begin{array}{l}x=2 x=- dfrac{3}{2} end{array} right. ) Vậy phương trình có tập nghiệm là : S={2;-(3)/(2)}