Toán Lớp 8: Tìm x biết a/ $x^{2}$ -8x+9=0 b/ $(3x-4)^{2}$ = $(x+2)^{2}$ c/x(x-2021)-x=-2021

Question

Toán Lớp 8:  Tìm x biết a/ $x^{2}$ -8x+9=0 b/ $(3x-4)^{2}$ =  $(x+2)^{2}$  c/x(x-2021)-x=-2021

thanhtung · Answer

Giải đáp:     a. S={&plusmn;4 sqrt{7}}   b. S={3; frac{1}{2}}   c. S={2021;1}     Lời giải và giải thích chi tiết:     a. x^2-8x+9=0   &harr;x^2-2.x.4+4^2-4^2+9=0   &harr;(x-4)^2-7=0   &harr; ( left[  begin{array}{l}x-4- sqrt{7}=0  x-4+ sqrt{7}=0 end{array}  right. )    &harr; ( left[  begin{array}{l}x=4+ sqrt{7}  x=4- sqrt{7} end{array}  right. )    Vậy: S={4+ sqrt{7};4- sqrt{7}}   b. (3x-4)^2=(x+2)^2   &harr;(3x-4)^2-(x+2)^2=0   &harr;(3x-4-x-2)(3x-4+x+2)=0   &harr;(2x-6)(4x-2)=0   &harr; ( left[  begin{array}{l}2x-6=0  4x-2=0 end{array}  right. )    &harr; ( left[  begin{array}{l}x=3  x= dfrac{1}{2} end{array}  right. )    Vậy: S={3; frac{1}{2}}   c. x(x-2021)-x=-2021   &harr;x(x-2021)-x+2021=0   &harr;x(x-2021)-(x-2021)=0   &harr;(x-2021)(x-1)=0   &harr; ( left[  begin{array}{l}x=2021  x=1 end{array}  right. )    Vậy: S={2021;1}

tuyethutanhthu · Answer

Giải đáp:       Lời giải và giải thích chi tiết:    a) x^2-8x+9=0   &hArr; x^2-8x+16-7=0   &hArr; (x-4)^2-7=0   &hArr; (x-4)^2-( sqrt{7})^2=0   &hArr; (x-4- sqrt{7})(x-4+ sqrt{7})=0   &hArr;  ( left[  begin{array}{l}x-4- sqrt{7}=0  x-4+ sqrt{7}=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=4+ sqrt{7}  x=4- sqrt{7} end{array}  right. )    Vậy S={4+ sqrt{7};4- sqrt{7}}   b) (3x-4)^2=(x+2)^2   &hArr; (3x-4)^2-(x+2)^2=0   &hArr; (3x-4-x-2)(3x-4+x+2)=0   &hArr; (2x-6)(4x-2)=0   &hArr;  ( left[  begin{array}{l}2x-6=0  4x-2=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=3  x= dfrac{1}{2} end{array}  right. )    Vậy S={3;1/2}   c) x(x-2021)-x=-2021   &hArr; x(x-2021)-x+2021=0   &hArr; x(x-2021)-(x-2021)=0   &hArr;(x-1)(x-2021)=0   &hArr;  ( left[  begin{array}{l}x-1=0  x-2021=0 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=1  x=2021 end{array}  right. )    Vậy S={1;2021}