Toán Lớp 8: B1: giải các pt sau: a,(3x+1)^2-(2x-5)^2=0 b,(x+6)(3x+1)+x^2-36=0 c,(x+3)(4-3x)=x^2+6x+9 d,(4x+3)^2-4(x-1)^2=0 e,(x+5)^2(3x+2)^2=x^2(x+

Question

Toán Lớp 8:  B1: giải các pt sau: a,(3x+1)^2-(2x-5)^2=0 b,(x+6)(3x+1)+x^2-36=0 c,(x+3)(4-3x)=x^2+6x+9 d,(4x+3)^2-4(x-1)^2=0 e,(x+5)^2(3x+2)^2=x^2(x+5)^2 f,(2x-1)(x-3)^2=(2x+1)(2x-1)^2 làm hộ mik vs ,mik cần gấp!!

ngoclanquynh · Answer

a) (3x+1)^2-(2x-5)^2=0 <=>(3x+1-2x+5)(3x+1+2x-5)=0 <=>(x+6)(5x-4)=0 <=> ( left[ begin{array}{l}x+6=0 5x-4=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-6 x= dfrac{4}{5} end{array} right. ) Vậy phương trình có tập nghiệm là S={-6;4/5} $ $ b) (x+6)(3x+1)+x^2-36=0 <=>(x+6)(3x+1)+(x-6)(x+6)=0 <=>(x+6)(3x+1+x-6)=0 <=>(x+6)(4x-5)=0 <=> ( left[ begin{array}{l}x+6=0 4x-5=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-6 x= dfrac{5}{4} end{array} right. ) Vậy phương trình có tập nghiệm là S={-6;5/4} $ $ c) (x+3)(4-3x)=x^2+6x+9 <=>(x+3)(4-3x)=(x+3)^2 <=>(x+3)(4-3x)-(x+3)^2=0 <=>(x+3)(4-3x-x-3)=0 <=>(x+3)(1-4x)=0 <=> ( left[ begin{array}{l}x+3=0 1-4x=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x= dfrac{1}{4} end{array} right. ) Vậy phương trình có tập nghiệm là S={-3;1/4} $ $ d) (4x+3)^2-4(x-1)^2=0 <=>(4x+3)^2-[2(x-1)]^2=0 <=>(4x+3)^2-(2x-2)^2=0 <=>(4x+3-2x+2)(4x+3+2x-2)=0 <=>(2x+5)(6x+1)=0 <=> ( left[ begin{array}{l}2x+5=0 6x+1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=- dfrac{5}{2} x=- dfrac{1}{6} end{array} right. ) Vậy phương trình có tập nghiệm là S={-5/2;-1/6} $ $ e) (x+5)^2(3x+2)^2=x^2(x+5)^2 <=>[(x+5)(3x+2)]^2=[x(x+5)]^2 <=>(3x^2+17x+10)^2=(x^2+5x)^2 <=>(3x^2+17x+10-x^2-5x)(3x^2+17x+10+x^2+5x)=0 <=>(2x^2+12x+10)(4x^2+22x+10)=0 <=>[2(x+1)(x+5)].[2(2x+1)(x+5)]=0 <=>(x+1)(x+5)^2(2x+1)=0 <=> ( left[ begin{array}{l}x+1=0 (x+5)^2=0 2x+1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-1 x=-5 x=- dfrac{1}{2} end{array} right. ) Vậy phương trình có tập nghiệm là S={-1;-5;-1/2} $ $ f) (2x-1)(x-3)^2=(2x+1)(2x-1)^2 <=>(2x-1)(x-3)^2-(2x+1)(2x-1)^2=0 <=>(2x-1)[(x-3)^2-(2x+1)(2x-1)]=0 <=>(2x-1)(x^2-6x+9-4x^2+1)=0 <=>(2x-1)(-3x^2-6x+10)=0 <=> ( left[ begin{array}{l}2x-1=0 -3x^2-6x+10=0 end{array} right. ) <=> ( left[ begin{array}{l}x= dfrac{1}{2} x=-1- dfrac{1}{3} sqrt{39} x=-1+ dfrac{1}{3} sqrt{39} end{array} right. ) Vậy phương trình có tập nghiệm S={1/2;-1+- sqrt{39}}

haiyemy · Answer

a)   (3x+1)^2-(2x-5)^2=0   &hArr;(3x+1-2x+5)(3x+1+2x-5)=0   &hArr;(x+6)(5x-4)=0   1)x+6=0&hArr;x=-6   2)5x-4=0&hArr;x=4/5   Vậy S={-6;4/5}   b)   (x+6)(3x+1)+x^2-36=0   &hArr;(x+6)(3x+1)+(x^2-6^2)=0   &hArr;(x+6)(3x+1)+(x-6)(x+6)=0   &hArr;(x+6)(3x+1+x-6)=0   &hArr;(x+6)(4x-5)=0   1)x+6=0&hArr;x=-6   2)4x-5=0&hArr;x=5/4   Vậy S={-6;5/4}   c)   (x+3)(4-3x)=x^2+6x+9   &hArr;(x+3)(4-3x)=(x+3)^2   &hArr;(x+3)(4-3x)-(x+3)^2=0   &hArr;(x+3)(4-3x-x-3)=0   &hArr;(x+3)(-4x+1)=0   1)x+3=0&hArr;x=-3   2)-4x+1=0&hArr;x=1/4   Vậy S={-3;1/4}   d)(4x+3)^2-4(x-1)^2=0   &hArr;(4x+3)^2-[2(x-1)]^2=0   &hArr;[4x+3-2(x-1)].[4x+3+2(x-1)]=0   &hArr;(4x+3-2x+2)(4x+3+2x-2)=0   &hArr;(2x+5)(6x+1)=0   1)2x+5=0&hArr;x=-5/2   2)6x+1=0&hArr;x=-1/6   Vậy S={-5/2;-1/6}   e)(x+5)^2(3x+2)^2=x^2(x+5)^2   &hArr;(x+5)^2(3x+2)^2-x^2(x+5)^2=0   &hArr;(x+5)^2[(3x+2)^2-x^2]=0   &hArr;(x+5)^2(3x+2-x)(3x+2+x)=0   &hArr;(x+5)^2(2x+2)(4x+2)=0   1)(x+5)^2=0&hArr;x+5=0&hArr;x=-5   2)2x+2=0&hArr;x=-1   3)4x+2=0&hArr;x=-1/2   Vậy S={-5;-1;-1/2}   f)(2x-1)(x-3)^2=(2x+1)(2x-1)^2   &hArr;(2x-1)(x-3)^2-(2x+1)(2x-1)^2=0   &hArr;(2x-1)[(x-3)^2-(2x+1)(2x-1)]=0   &hArr;(2x-1)(x^2-6x+9-4x^2+1)=0   &hArr;(2x-1)(-3x^2-6x+10)=0   1)2x-1=0&hArr;x=1/2   2)-3x^2-6x+10=0&hArr;x=+-$ dfrac{ sqrt[]{39} }{3}$-1    Vậy S={1/2; +-$ dfrac{ sqrt[]{39} }{3}-1$}