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Toán Lớp 8: tìm x biết 1, 2021x(x-2022)-x+2022=0 2,(2x+1)^2=(x-2)^2

Home/ Toán Lớp 8: tìm x biết 1, 2021x(x-2022)-x+2022=0 2,(2x+1)^2=(x-2)^2

Toán Lớp 8: tìm x biết 1, 2021x(x-2022)-x+2022=0 2,(2x+1)^2=(x-2)^2

Nhiên Tháng Mười 28, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x biết
1, 2021x(x-2022)-x+2022=0
2,(2x+1)^2=(x-2)^2

Comments ( 2 )

hiennhi98
28 Tháng Mười, 2022 at 10:01 sáng

Reply

1.

2021x.(x-2022)-x+2022=0

2021x.(x-2022)-(x-2022)=0

(x-2022)(2021x-1)=0

TH1

x-2022=0

x=2022

TH2

2021x-1=0

2021x=1

x=1/2021

2.

(2x+1)^2=(x-2)^2

(2x+1)^2-(x-2)^2=0

(2x+1-x+2)(2x+1+x-2)=0

(x+3)(3x-1)=0

TH1

x+3=0

x=-3

TH2

3x-1=0

3x=1

x=1/3
minhtamnguyet88
28 Tháng Mười, 2022 at 10:01 sáng

Reply

2021x(x-2022)-x+2022=0

⇔2021x(x-2022)-(x-2022)=0

⇔(x-2022)(2021x-1)=0

⇔x-2022=0⇔x=2022

2021x-1=0⇔2021x=1⇔x=1/2021

Vậy x={2022,1/2021}

b,

(2x+1)^2=(x-2)^2

⇔(2x+1)^2-(x-2)^2=0

⇔(2x+1-x+2)(2x+1+x-2)=0

⇔(x+3)(3x-1)=0

⇔x+3=0⇔x=-3

3x-1=0⇔3x=1⇔x=1/3

Vậy x={-3,1/3}

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