\(BĐT\Leftrightarrow a^2+5b^2-3a-b-3ab+5\ge0\)

           \(\Leftrightarrow2a^2+10b^2-6a-2b-6ab+10\ge0\)

           \(\Leftrightarrow\left(a^2-6ab+9b^2\right)+\left(a^2-2a+9\right)+\left(b^2-2b+1\right)\ge0\)

           \(\Leftrightarrow\left(a-36b\right)^2+\left(a-3\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng với mọi a; b)

Vậy bất đẳng thức được chứng minh.Đẳng thức xảy ra <=> (a; b) = (3; 1)

Ta có: \(a^2-b^2=4c^2\)

\(\Rightarrow a^2-b^2-4c^2=0\)

Xét hiệu:

 \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2\)

\(=\left(5a-3b\right)^2-\left(8c\right)^2-\left(3a-5b\right)^2\)

\(=25a^2-30ab+9b^2-64c^2-9a^2+30ab-25b^2\)

\(=16a^2-16b^2-64c^2\)

\(=16\left(a^2-b^2-4c^2\right)\)

\(=16.0\)

\(=0\)

\(\Rightarrow\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)

                                                                             đpcm 

Tham khảo nhé~

Một cách khác :))

Xét VT của biểu thức cần cm ta có :

( 5a - 3b + 8c )( 5a - 3b - 8c )

= [ ( 5a - 3b ) + 8c ][ ( 5a - 3b ) - 8c ]

= ( 5a - 3b )² - ( 8c )²

= 25a² - 30ab + 9b² - 64c²

= 25a² - 30ab + 9b² - 16.4c²

= 25a² - 30ab + 9b² - 16( a² - b² ) < theo đề a² - b² = 4c² >

= 25² - 30ab + 9b² - 16a² + 16b²

= 9a² - 30ab + 25b²

= ( 3a - 5b )² = VP

=> đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

a: \(\frac{3a+5b}{2a-7b}=\frac{3\cdot bk+5b}{2\cdot bk-7b}=\frac{b\left(3k+5\right)}{b\left(2k-7\right)}=\frac{3k+5}{2k-7}\)

\(\frac{3c+5d}{2c-7d}=\frac{3\cdot dk+5d}{2\cdot dk-7d}=\frac{d\left(3k+5\right)}{d\left(2k-7\right)}=\frac{3k+5}{2k-7}\)

Do đó: \(\frac{3a+5b}{2a-7b}=\frac{3c+5c}{2c-7d}\)

b: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)

\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)

Do đó: \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)

Bạn kiểm tra lại đề nhé.

G/s: x = y \(\ne\)0 => a = b 

=> \(2a^2.2x^2=4a^2\) ???

ta có : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)

\(\Leftrightarrow\left(5a-3b\right)^2-\left(8c\right)^2=\left(3a-5b^2\right)\)

\(\Leftrightarrow\left(5a-3b\right)^2-\left(3a-5b\right)^2=\left(8c\right)^2\)

\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=\left(8c\right)^2\)

\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=64c^2\)

\(\Leftrightarrow16\left(a^2-b^2\right)=64c^2\Leftrightarrow a^2-b^2=4c^2\) đúng như giả thiết

\(\Rightarrow\left(đpcm\right)\)

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(5a-3b\right)^2-64c^2=25a^2-2.5.3ab+9b^2-16\left(a^2-b^2\right)\)

\(=9a^2-2.3.5ab+25b^2=\left(3a-5b\right)^2\)