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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: |x+3| + |x-y| =0 |x-1| + (y-2)2 =0 |x-3| + |x-2y+1| =0 |x-2| + |3x .y-5| =0 |x-2| – 3|2-x| =-10 |x-3|+ |x-y+1|=0 |x-1| + (y+5)2020 =0

Home/ Toán Lớp 7: |x+3| + |x-y| =0 |x-1| + (y-2)2 =0 |x-3| + |x-2y+1| =0 |x-2| + |3x .y-5| =0 |x-2| – 3|2-x| =-10 |x-3|+ |x-y+1|=0 |x-1| + (y+5)2020 =0

Toán Lớp 7: |x+3| + |x-y| =0 |x-1| + (y-2)2 =0 |x-3| + |x-2y+1| =0 |x-2| + |3x .y-5| =0 |x-2| – 3|2-x| =-10 |x-3|+ |x-y+1|=0 |x-1| + (y+5)2020 =0

Tháng Hai 4, 2023 Toán học 2 Comments 0 views

Toán Lớp 7: |x+3| + |x-y| =0
|x-1| + (y-2)2 =0
|x-3| + |x-2y+1| =0
|x-2| + |3x .y-5| =0
|x-2| – 3|2-x| =-10
|x-3|+ |x-y+1|=0
|x-1| + (y+5)2020 =0

Comments ( 2 )

  1. annhocbichchaubich
    annhocbichchaubich
    4 Tháng Hai, 2023 at 1:04 sáng
    Reply
    a) |x+3| + |x-y| =0
    Do |x+3| ≥ 0; |x-y| ≥ 0
    => |x+3| + |x-y| ≥ 0
    Dấu “=” có khi:
    |x+3| = 0; |x-y| = 0
    => x+3 = 0; |x-y| = 0
    => x = -3; x-y = 0
    => x = -3; -3 -y = 0
    => x = -3; y = -3
    Vậy x = -3; y = -3
    b) |x-1| + (y-2)^2 =0
    Do |x-1| ≥ 0; (y-2)^2 ≥ 0
    => |x-1| + (y-2)^2 ≥ 0
    Dấu “=” có khi:
    |x-1| = 0; (y-2)^2 = 0
    => x – 1 = 0; y – 2 = 0
    => x = 1; y = 2
    Vậy x = 1; y = 2
    c) |x-3| + |x-2y+1| =0
    Do |x-3| ≥ 0; |x-2y+1| ≥ 0
    => |x-3| + |x-2y+1| ≥ 0
    Dấu “=” có khi:
    |x-3| =0; |x-2y+1| =0
    => x-3 =0; |x-2y+1| =0
    => x = 3; 3-2y+1 = 0
    => x = 3; 2 – 2y = 0
    => x = 3; 2y = 2
    => x = 3; y = 1
    Vậy x = 3; y = 1
    d) |x-2| + |3x .y-5| =0
    Do |x-2| ≥ 0; |3x .y-5| ≥ 0
    => |x-2| + |3x .y-5| ≥ 0
    Dấu “=” có khi:
    |x-2| = 0; |3x .y-5| = 0
    => x-2 = 0; |3x .y-5| = 0
    => x = 2; 3 . 2 .y-5 = 0
    => x = 2; 6y-5 = 0
    => x = 2; 6y = 5
    => x = 2; y = 5/6
    Vậy x = 2; y = 5/6
    e) |x-2| – 3|2-x| =-10
    => 1|x-2| – 3|2-x| =-10
    Ta có: |x-2| = |2-x|
    => (1-3)|x-2| = -10
    => -2|x-2| = -10
    => |x-2| = 5
    Th1: x-2 = 5
    => x = 7
    Th1: x-2 = -5
    => x = -3
    Vậy x = -3 hoặc 7
    g) |x-3|+ |x-y+1|=0
    Do |x-3| ≥ 0;  |x-y+1| ≥ 0
    => |x-3|+ |x-y+1| ≥ 0
    Dấu “=” có khi:
    |x-3| ≥ 0;  |x-y+1| ≥ 0
    => x-3 = 0;  x-y+1 = 0
    => x = 3; 3-y+1 = 0
    => x = 3; 4-y = 0
    => x = 3; y=4
    Vậy x = 3; y=4
    h) |x-1| + (y+5)^2020 =0
    Do |x-1| ≥ 0; (y+5)^2020 ≥ 0
    => |x-1| + (y+5)^2020 ≥ 0
    Dấu “=” có khi:
    |x-1| = 0; (y+5)^2020 = 0
    => x – 1 = 0; y + 5 = 0
    => x = 1; y = -5
    Vậy x = 1; y = -5
    (Chúc bạn học tốt)

  2. giahan44
    giahan44
    4 Tháng Hai, 2023 at 1:04 sáng
    Reply
    Giải đáp:
    |x+3|+|x-y|=0
    Vì |x+3|>=0,|x-y|>=0
    =>|x-3|+|x-y|=0
    Dấu “=” xảy ra khi \(\begin{cases}x=3\\y=x=3\\\end{cases}\)
    |x-1|+(y-2)^2=0
    Vì |x-1|>=0,(y-2)^2>=0
    =>|x-1|+(y-2)^2>=0
    Dấu “=” xảy ra khi \(\begin{cases}x=1\\y=2\\\end{cases}\)
    |x-3|+|x-2y+1|=0
    Vì |x-3|>=0,|x-2y+1|>=0
    =>|x-3|+|x-2y+1|>=0
    Dấu “=” xảy ra khi \(\begin{cases}x=3\\y=\dfrac{x+1}{2}=2\\\end{cases}\)
    |x-2|+|3xy-5|=0
    Vì |x-2|>=0,|3xy-5|>=0
    =>|x-2|+|3xy-5|=0
    Dấu “=” xảy ra khi \(\begin{cases}x=2\\y=\dfrac56\\\end{cases}\)
    |x-2|-3|2-x|=-10
    <=>-|x-2|-3|x-2|=-10
    <=>-2|x-2|=-10
    <=>|x-2|=5
    <=> \(\left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 
    |x-3|+|x-y+1|=0
    Vì |x-3|>=0
    |x-y+1|>=0
    =>|x-3|+|x-y+1|=0
    Dấu “=” xảy ra khi \(\begin{cases}x=3\\y=x+1=4\\\end{cases}\)
    |x-1|+(y+5)^2020=0
    Vì |x-1|>=0
    (y+5)^2020>=0
    =>|x-1|+(y+5)^2020>=0
    Dấu “=” xảy ra khi \(\begin{cases}x=1\\y=-5\\\end{cases}\)

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222-9+11+12:2*14+14 = ? ( )

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