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

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

\(\Rightarrow\left(đpcm\right)\)

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

      \(\left(5a-3b+4c\right)\left(5a-3b-4c\right)\)

\(=\left(5a-3b\right)^2-\left(4c\right)^2\)

\(=25a^2-30ab+9b^2-16c^2\)

\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)

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

\(=\left(3a\right)^2-2.3a.5b+\left(5b\right)^2=\left(3a-5b\right)^2\)

Chúc bạn học tốt.

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

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

xét hiệu\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2=0\)

\(\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)

\(\left(5a-3b\right)^2-\left(3a-5b\right)^2-64c^2=0\)

\(\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)-64c^2=0\)

\(\left(2a+2b\right)\left(8a-8b\right)-64c^2=0\)

\(16a^2-16ab+16ab-16b^2-64c^2=0\)

\(16a^2-16b^2-64c^2=0\)

\(16\left(a^2-b^2\right)-64c^2=0\)

\(16\times4c^2-64c^2=0\)

\(64c^2-64c^2=0\left(dpcm\right)\)

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

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

\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)

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

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

( 5a - 3b + 8c )( 5a - 3b - 8c ) 

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

= ( 5a - 3b )² - ( 8c )²

= 25a² - 30ab + 9c² - 64c²

= 25a² - 30ab + 9c² - 16.4c²

= 25a² - 30ab + 9c² - 16( a² - b² )

= 25a² - 30ab + 9c² - 16a² + 16b²

= 9a² - 30ab + 25b²

= ( 3a - 5b )² ( đpcm )

\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)=\left(3a-5b\right)^2\)

\(\Leftrightarrow25a^2-15ab-20ac-15ab+9b^2+12bc+20ac-12bc-16c^2=9a^2-30ab+25b^2\)

\(\Leftrightarrow25a^2-30ab+9b^2-16c^2=9a^2-30ab+25b^2\)

\(\Leftrightarrow25a^2+9b^2-16c^2=9a^2+25b^2\)

\(\Leftrightarrow16a^2=16c^2+16b^2\)

\(\Rightarrow a^2=b^2+c^2\)

\(\Rightarrow\Delta\) với 3 cạnh a, b, c vuông

\(\Rightarrow\Delta\) có độ dài 3 cạnh trên là \(\Delta\) vuông ( đpcm )

Vậy...

Ta có:

sigma \(\frac{ab}{3a+4b+5c}=\) sigma \(\frac{2ab}{5\left(a+b+2c\right)+\left(a+3b\right)}\le\frac{2}{36}\left(sigma\frac{5ab}{a+b+2c}+sigma\frac{ab}{a+3b}\right)\)

Ta đi chứng minh: \(sigma\frac{ab}{a+b+2c}\le\frac{9}{4}\)

có: \(sigma\frac{ab}{a+b+2c}\le\frac{1}{4}\left(sigma\frac{ab}{c+a}+sigma\frac{ab}{b+c}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

BĐT trên đúng nếu: \(sigma\frac{ab}{a+3b}\le\frac{9}{4}\)

Ta thấy: \(sigma\frac{ab}{a+3b}\le\frac{1}{16}\left(sigma\frac{ab}{a}+sigma\frac{3ab}{b}\right)=\frac{1}{16}\)( sigma \(b+sigma3a\)) \(=\frac{1}{4}\left(a+b+c\right)=\frac{9}{4}\)

\(\Leftrightarrow sigma\frac{ab}{3a+4b+5c}\le\frac{1}{18}\left(5.\frac{9}{4}+\frac{9}{4}\right)=\frac{3}{4}\)(1)

MÀ: \(\frac{1}{\sqrt{ab\left(a+2c\right)\left(b+2c\right)}}=\frac{2}{2\sqrt{\left(ab+2bc\right)\left(ab+2ca\right)}}\ge\frac{2}{2\left(ab+bc+ca\right)}\)

\(=\frac{3}{3\left(ab+bc+ca\right)}\ge\frac{3}{\left(a+b+c\right)^2}=\frac{3}{9^2}=\frac{1}{27}\)(2)

Từ (1) và (2) \(\Rightarrow T\le\frac{3}{4}-\frac{1}{27}=\frac{77}{108}\)

Vậy GTLN của biểu thức T là 77/108 <=> a=b=c=3