Toán Lớp 8: Giải các phương trình sau 1, (x+3)^2 ×(x-4)=0 2, (2x-4)^2 ×(2x-6)=0 3, (4x+8)^2 ×(2x-6)=0 4, (x^2+1)×(x-1)^2×(3x-1)=0

Question

Toán Lớp 8:  Giải các phương trình sau 1,  (x+3)^2 ×(x-4)=0 2,   (2x-4)^2 ×(2x-6)=0 3,    (4x+8)^2 ×(2x-6)=0 4,     (x^2+1)×(x-1)^2×(3x-1)=0

kimlien · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    1.  (x+3)&sup2; &times;(x-4)=0    &hArr;  ( left[  begin{array}{l}(x+3)&sup2;=0  x-4=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=-3  x=4 end{array}  right. )    Vậy tập nghiệm pt S={-3;4}    2.  (2x-4)&sup2; &times;(2x-6)=0     &hArr;  ( left[  begin{array}{l}(2x-4)&sup2;=0  2x-6=0 end{array}  right. )      &hArr;  ( left[  begin{array}{l}x=2  x=3 end{array}  right. )     Vậy  tập nghiệm pt S={2;3}    3.  (4x+8)&sup2; &times;(2x-6)=0     &hArr;  ( left[  begin{array}{l}(4x+8)&sup2;=0  2x-6=0 end{array}  right. )      &hArr;  ( left[  begin{array}{l}x=-2  x=3 end{array}  right. )      Vậy  tập nghiệm pt S={-2;3}      4. (x&sup2;+1)&times;(x-1)&sup2;&times;(3x-1)=0     &hArr; x&sup2;+1=0 hoặc  ( left[  begin{array}{l}(x-1)&sup2;=0  3x-1=0 end{array}  right. )      &hArr; x&isin;&empty;  hoặc  ( left[  begin{array}{l}x=1  x= frac{1}{3} end{array}  right. )        Vậy  tập nghiệm pt S={1;$ frac{1}{3}$ }       -Học tốt-

lyly98 · Answer

$ text {1,}$     (x+3)^2.(x-4) =0     &hArr; ( left[  begin{array}{l}(x+3)^2=0  x-4=0 end{array}  right. )      &hArr; ( left[  begin{array}{l}x+3=0  x=4 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=-3  x=4 end{array}  right. )      $ text {Vậy S&isin;}$ {-3;4}     $ text {2,}$     (2x-4)^2.(2x-6) =0     &hArr; ( left[  begin{array}{l}(2x-4)^2=0  2x-6 =0 end{array}  right. )      &hArr; ( left[  begin{array}{l}2x-4 =0  2x=6 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=2  x=3 end{array}  right. )      $ text {Vậy S&isin;}$ {2;3}     $ text {3,}$     (4x+8)^2.(2x-6) =0     &hArr; ( left[  begin{array}{l}(4x+8)^2 =0  2x-6 =0 end{array}  right. )      &hArr; ( left[  begin{array}{l}4x+8=0  2x=6 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=-2  x=3 end{array}  right. )      $ text {Vậy S&isin;}$ {-2;3}     $ text {4,}$     (x^2+1)(x-1)^2(3x-1) =0     &hArr; ( left[  begin{array}{l}x^2+1=0  (x-1)^2=0  3x-1=0 end{array}  right. )      &hArr; ( left[  begin{array}{l}x^2=-1 (Kh&ocirc;ng TM)  x-1=0  3x=1 end{array}  right. )      &hArr; ( left[  begin{array}{l}x^2=-1 (Kh&ocirc;ng TM)  x=1  x= frac{1}{3}  end{array}  right. )      $ text {Vậy S&isin;}$ {1; frac{1}{3}}      $ text {@lamtung2007}$