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

Toán Lớp 7: Cho $\frac{a}{b}$ = $\frac{c}{d}$ . CMR : a ) $\frac{ab}{cd}$ = $\frac{a^{2}- b^2}{c^{2}- d^2}$ b ) $\frac{ab}{cd}$ = $\frac{a^{2

Toán Lớp 7: Cho $\frac{a}{b}$ = $\frac{c}{d}$ . CMR :
a ) $\frac{ab}{cd}$ = $\frac{a^{2}- b^2}{c^{2}- d^2}$
b ) $\frac{ab}{cd}$ = $\frac{a^{2}+ b^2}{c^{2}+ d^2}$
c ) $\frac{ab}{cd}$ = $\frac{(a- b)^2}{(c- d)^2}$
f

Comments ( 2 )

  1. Trả lời:
    $\text{Đặt}$ a/b=c/d=k;a=bk;c=dk
    $\text{Ta có:}$
    a,(ab)/(cd)=(bkb)/(dkd)=(b^2)/(d^2)(1)
    (a^2-b^2)/(c^2-d^2)=((bk)^2-b^2)/((dk)^2-d^2)=(b^2 .(k^2-1))/(d^2 .(k^2-1))=(b^2)/(d^2)(2)
    $\text{Từ}$ (1);(2)=>(ab)/(cd)=(a^2-b^2)/(c^2-d^2)
    b,(ab)/(cd)=(bkb)/(dkd)=(b^2)/(d^2)(1)
    (a^2+b^2)/(c^2+d^2)=((bk)^2+b^2)/((dk)^2+d^2)=(b^2 .(k^2+1))/(d^2 .(k^2+1))=(b^2)/(d^2)(2)
    $\text{Từ}$ (1);(2)=>(ab)/(cd)=(a^2+b^2)/(c^2+d^2)
    c,(ab)/(cd)=(bkb)/(dkd)=(b^2)/(d^2)(1)
    ((a-b)^2)/((c-d)^2)=((bk-b)^2)/((dk-d)^2)=([b.(k-1)]^2)/([d.(k-1)]^2)=(b^2 .(k-1)^2)/(d^2 .(k-1)^2)=(b^2)/(d^2)(2)
    $\text{Từ}$ (1);(2)=>(ab)/(cd)=((a-b)^2)/((c+d)^2)

  2. Giải đáp:
    Đặt ab=cd=k;a=bk;c=dk
    Ta có:
    a,abcd=bkbdkd=b2d2(1)
    a2-b2c2-d2=(bk)2-b2(dk)2-d2=b2.(k2-1)d2.(k2-1)=b2d2(2)
    Từ (1);(2)⇒abcd=a2-b2c2-d2
    b,abcd=bkbdkd=b2d2(1)
    a2+b2c2+d2=(bk)2+b2(dk)2+d2=b2.(k2+1)d2.(k2+1)=b2d2(2)
    Từ (1);(2)⇒abcd=a2+b2c2+d2
    c,abcd=bkbdkd=b2d2(1)
    (a-b)2(c-d)2=(bk-b)2(dk-d)2=[b.(k-1)]2[d.(k-1)]2=b2.(k-1)2d2.(k-1)2=b2d2(2)
    Từ 
    Lời giải và giải thích chi tiết:
     

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