a)  \(x^2+10x+25+y^2+1+2y=\left(x+5\right)^2+\left(y+1\right)^2\)

b)  \(x^2-2xy+2y^2+2y+1=\left(x-y\right)^2+\left(y+1\right)^2\)

c)  \(x^2-6x+13+y^2+4y=\left(x-3\right)^2+\left(y+2\right)^2\)

d) \(4x^2+2y^2-4xy-2y+1=\left(2x-y\right)^2+\left(y-1\right)^2\)

❤ѕѕѕσиɢσкυѕѕѕ❤

Ta có: \(x^2+y^2+z^2+t^2-\left(xy+yz+zt+tx\right)=1-1\)

\(\Leftrightarrow2\left(x^2+y^2+z^2+t^2-xy-yz-zt-tx\right)=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2+2t^2-2xy-2yz-2zt-tx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zt+t^2\right)+\left(t^2-2tx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2=0\)

Vì \(\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0;\left(z-t\right)^2\ge0;\left(t-x\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2\ge0\)

Dấu "=" xảy ra khi x - y = 0 ; y - z = 0 ; z - t = 0 ; t - x = 0 <=> x = y = z = t

Khi đó \(x^2+y^2+z^2+t^2=x^2+x^2+x^2+x^2=4x^2=1\)

\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)

Vậy \(x=y=z=t=\pm\frac{1}{2}\)

༺Ɗเευ༒Ƭɦυyεɳ༻

❤ѕѕѕσиɢσкυѕѕѕ❤

Thế b²=ad vào ta được: \(\frac{a^2+b^2}{b^2+d^2}=\frac{a^2+ad}{ad+d^2}=\frac{a\left(a+d\right)}{d\left(a+d\right)}=\frac{a}{d}\left(đpcm\right)\)

Ta có : 

\(b^2=ad\)

\(\Rightarrow\frac{a}{b}=\frac{b}{d}\)

\(\Rightarrow\frac{a}{b}.\frac{a}{b}=\frac{b}{d}.\frac{b}{d}=\frac{a}{b}.\frac{b}{d}\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{d^2}=\frac{a}{d}\)

ADTCDTSBN , ta có : 

\(\frac{a^2}{b^2}=\frac{b^2}{d^2}=\frac{a}{d}=\frac{a^2+b^2}{b^2+d^2}=\frac{a}{d}\)

\(\Rightarrow\frac{a^2+b^2}{b^2+d^2}=\frac{a}{d}\left(đpcm\right)\)

Ta có : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

1) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{b}{d}=\frac{a}{c}=\frac{a+b}{c+d}\)

\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)

                         đpcm

2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)

\(\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

                         đpcm