Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)

Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)

c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)

\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)

Cộng x với z

ra HĐT suy ra 

\(x+z=\left(a-b\right)^2+\left(c-d\right)^2+a^2+c^2\)

do a,b,c,d>0 nên x+z>0 vậy 1 trong 2 số có ít nhất 1 số dương 

tương tự tự làm nhé

cảm ơn nhé

a) Ta có: \(a^2+2ab+b^2-c^2+2cd-d^2\)

\(=\left(a^2+2ab+b^2\right)-\left(c^2-2cd+d^2\right)\)

\(=\left(a+b\right)^2-\left(c-d\right)^2\)

\(=\left(a+b-c+d\right)\left(a+b+c-d\right)\)

b) Ta có: \(x^2-4xy+4y^2-x+2y\)

\(=\left(x-2y\right)^2-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x-2y-1\right)\)

\(VT=a^2+b^2+c^2+d^2-2\left(a+c\right)\left(b+d\right)\)

\(VT\ge\frac{1}{4}\left(a+b+c+d\right)^2-\frac{1}{2}\left(a+b+c+d\right)^2=-\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=d=\frac{1}{4}\)

a² + b² - c² - d² + 2cd - 2ab

= (a² - 2ab + b²) - (c² -2cd + d²)

= (a-b)² - (c-d)²

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

\(\left(a-b\right)^2-\left(c-d\right)^2=\left[\left(a-b\right)-\left(c-d\right)\right]\left[\left(a-b\right)+c-d\right]\)

\(\left(a+d-b-c\right)\left(a+c-b-d\right)\)