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

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1. Giải đáp:

   ĐặtĐặt ab=cd=k;a=bk;c=dkab=cd=k;a=bk;c=dk

   Ta có:Ta có:

   a,abcd=bkbdkd=b2d2(1)a,abcd=bkbdkd=b2d2(1)

   a2−b2c2−d2=(bk)2−b2(dk)2−d2=b2.(k2−1)d2.(k2−1)=b2d2(2)a2-b2c2-d2=(bk)2-b2(dk)2-d2=b2.(k2-1)d2.(k2-1)=b2d2(2)

   TừTừ (1);(2)⇒abcd=a2−b2c2−d2(1);(2)⇒abcd=a2-b2c2-d2

   b,abcd=bkbdkd=b2d2(1)b,abcd=bkbdkd=b2d2(1)

   a2+b2c2+d2=(bk)2+b2(dk)2+d2=b2.(k2+1)d2.(k2+1)=b2d2(2)a2+b2c2+d2=(bk)2+b2(dk)2+d2=b2.(k2+1)d2.(k2+1)=b2d2(2)

   TừTừ (1);(2)⇒abcd=a2+b2c2+d2(1);(2)⇒abcd=a2+b2c2+d2

   c,abcd=bkbdkd=b2d2(1)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)(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)

   TừTừ (1);(2)⇒abcd=(a−b)2(c+d)2

   Lời giải và giải thích chi tiết:

    

   Trả lời
2. 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)

   Trả lời