a)(a – b + c) – (a + c) = -b b) (a + b) – (b – a) + c = 2a + c c)- (a + b – c) + (a – b – c) = -2b d)a(b + c) – a(b + d) = a(c – d) e)a(b – c) + a(d +

a)(a – b + c) – (a + c) = -b
b) (a + b) – (b – a) + c = 2a + c
c)- (a + b – c) + (a – b – c) = -2b
d)a(b + c) – a(b + d) = a(c – d)
e)a(b – c) + a(d + c) = a(b + d)
f) a.(b – c) – a.(b + d) = -a.( c + d)
h) (a + b).( c + d) – (a + d).( b + c) = (a – c). (d – b)

0 bình luận về “a)(a – b + c) – (a + c) = -b b) (a + b) – (b – a) + c = 2a + c c)- (a + b – c) + (a – b – c) = -2b d)a(b + c) – a(b + d) = a(c – d) e)a(b – c) + a(d +”

  1. Đáp án:

    VT=Vế trái

    VP=Vế phải

    .

    `a)VT=(a-b+c)-(a+c)`

    `=a-b+c-a-c`

    `=(a-a)-b+(c-c)`

    `=-b=VP`

    `→VT=VP(đpcm)`

    `b)VT=(a+b)-(b-a)+c`

    `=a+b-b+a+c`

    `=(a+a)+(b-c)+c`

    `=2a+c=VP`

    `→VT=VP(đpcm)`

    `c)VT=-(a+b-c)+(a-b-c)`

    `=-a-b+c+a-b-c`

    `=(-a+a)+(-b-b)+(c-c)`

    `=-2b=VP`

    `→VT=VP(đpcm)`

    `d)VT=a(b+c)-a(b+d)`

    `=ab+ac-ab-ad`

    `=(ab-ab)+(ac-ad)`

    `=a(c-d)=VP`

    `→VT=VP(đpcm)`

    `e)VT=a(b-c)+a(d+c)`

    `=ab-ac+ad+ac`

    `=(ab+ad)+(-ac+ac)`

    `=a(b+d)=VP`

    `→VT=VP(đpcm)`

    `f)VT=a(b-c)-a(b+d)`

    `=ab-ac-ab-ad`

    `=(ab-ab)+(-ac-ad)`

    `=-a(c+d)=VP`

    `→VT=VP(đpcm)`

    `h)VT=(a+b)(c+d)-(a+d)(b+c)`

    `=ac+ad+bc+bd-(ab+ac+bd+cd)`

    `=ac+ad+bc+bd-ab-ac-bd-cd`

    `=ad+bc-ab-cd`

    `VP=(a-c)(d-b)`

    `=ad-ab-cd+bc`

    `→VT=VP(đpcm)`

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  2. Đáp án:

     

    Giải thích các bước giải:

     a)a-b+c-a-c+b=0 <=>0

    b)a+b-b+a+c-2a-c=0<=>0=0

    c)-a-b+c+a-b-c+2b<=>0=0

    d)ab+ac-ab-ad=ac-ad

    <=>ac-ad-ac+ad=0

    <=>0=0

    e) ab-ac+ad+ac=ab+ad

    =>ab+ad-ab-ad=0

    =>0=0

    f)ab-ac-ab-ad=-ac-ad

    =>-ac-ad+ac+ad=0

    =>0=0

    h)ac+ad+bc+bd-ab-ac-db-dc=ad=ab-cd+cb

    =>0=0

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