1. Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)
Thay vào 2 vế là sẽ CM được
1. Đặt \(\frac{a}{b}=\frac{c}{d}=k>a=bk.c=dk\)
Thay vào 2 vế để chứng minh
1 )
Ta có :
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)( Áp dụng t/c DTSBN )
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(1\right)\)
Lại có : \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{2b^2}{2d^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\) ( Áp dụng t/c DTSBN ) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\left(đpcm\right)\)
2 )
Ta có :
\(x+y+2xy=83\)
\(\Rightarrow2\left(x+y+2xy\right)=166\)
\(\Rightarrow2x+2y+4xy+1=167\)
\(\Rightarrow2x\left(2y+1\right)+\left(2y+1\right)=167\)
\(\Rightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Do \(x;y\in Z\)
\(\Leftrightarrow2x+1;2y+1\in Z\)
\(\Leftrightarrow2x+1;2y+1\in\left\{\pm1;\pm167\right\}\)
Ta có bảng sau :
\(2x+1\) | \(1\) | \(167\) | \(-1\) | \(-167\) |
\(2y+1\) | \(167\) | \(1\) | \(-167\) | \(-1\) |
\(x\) | \(0\) | \(83\) | \(-1\) | \(-84\) |
\(y\) | \(83\) | \(0\) | \(-84\) | \(-1\) |
Vậy \(\left(x;y\right)\in\left\{\left(0;83\right),\left(83;0\right),\left(-1;-84\right),\left(-84;-1\right)\right\}\)
Có a = bk
c = dk
Thay vào VT ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\frac{b}{d}\) (1)
VP = \(\frac{3.\left(bk\right)^2+2.b^2}{3.\left(dk\right)^2+2.d^2}=\frac{3.b^2.k^2+2.b^2}{3.d^2.k^2+2.d^2}=\frac{b^2\left(3.k^2+2\right)}{d^2\left(3.k^2+2\right)}=\frac{b}{d}\)(2)
Từ (1) và (2) => đpcm
Mình cũng không chắc lắm
TXT Channel Funfun :
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405