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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
\(\Rightarrow\left(đpcm\right)\)
xét hiệu\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-\left(3a-5b\right)^2-64c^2=0\)
\(\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)-64c^2=0\)
\(\left(2a+2b\right)\left(8a-8b\right)-64c^2=0\)
\(16a^2-16ab+16ab-16b^2-64c^2=0\)
\(16a^2-16b^2-64c^2=0\)
\(16\left(a^2-b^2\right)-64c^2=0\)
\(16\times4c^2-64c^2=0\)
\(64c^2-64c^2=0\left(dpcm\right)\)
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2\)
\(a^2-b^2-c^2=0\Rightarrow c^2=a^2-b^2\)
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)\)
\(=\left(5a-3b\right)^2-\left(4c\right)^2\)
\(=25a^2-30ab+9b^2-16c^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a\right)^2-2.3a.5b+\left(5b\right)^2=\left(3a-5b\right)^2\)
Chúc bạn học tốt.
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)=\left(3a-5b\right)^2\\ 25a^2-15ab-20ac-15ab+9b^2+12bc+20ac-12bc-16c^2=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2-30ab=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2=9a^2+25b^2\\ \Leftrightarrow25a^2-9a^2=-9b^2+25b^2+16c^2\\ \Leftrightarrow16a^2-=16b^2+16c^2\\ \Leftrightarrow a^2=b^2+c^2\)
Vậy ...
B1:
a)
\(A=11-10x-x^2\\ A=-x^2-10x-25+36\\ A=-\left(x-5\right)^2+36\le36\)
đẳng thức xảy ra khi x-5=0 => x=5
vậy GTLN của A là 36 tại x=5
b)
\(B=4-x^2+2x\\ B=-x^2+2x-1+5\\ B=-\left(x-1\right)^2+5\le5\)
đẳng thức xảy ra khi x-1=0 => x=1
c)
\(C=4x-x^2\\ C=-x^2+4x-4+4\\ C=-\left(x-2\right)^2+4\le4\)
đẳng thức xảy ra khi x-2=0 => x=2
Sửa đề: CMR : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
Bài 2:Ta có:
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
\(\Leftrightarrow\left(5a-3b\right)^2-64c^2=\left(3a-5b\right)^2\)
\(\Leftrightarrow\left(5a-3b\right)^2-\left(3a-5b\right)^2=64c^2\)
\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=64c^2\)
\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=64c^2\)
\(\Leftrightarrow16\left(a+b\right)\left(a-b\right)=64c^2\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)=4c^2\)
\(\Leftrightarrow a^2-b^2=4c^2\) ( Đúng )
\(\Rightarrow\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
\(\Leftrightarrow\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)
\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=64c^2\)
\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=16\left(a^2-b^2\right)\)
\(\Leftrightarrow16\left(a^2-b^2\right)=16\left(a^2-b^2\right)\left(true\right)\)
Vậy \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)khi \(a^2-b^2=4c^2\)
(5a−3b+8c)(5a−3b−8c)(5a-3b+8c)(5a-3b-8c)
=(5a−3b)2−(8c)2=(5a-3b)2-(8c)2
=(5a−3b)2−16.4c2=(5a-3b)2-16.4c2
Thay a2−b2=4c2a2-b2=4c2 ta có :
=25a2−30ab+9b2−16(a2−b2)=25a2-30ab+9b2-16(a2-b2)
=25a2−30ab+9b2−16a2+16b2=25a2-30ab+9b2-16a2+16b2
=9a2−30ab+25b2=9a2-30ab+25b2
=(3a−5b)2(đpcm)=(3a-5b)2(dpcm)
Ta có : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=\left(5a-3b\right)^2-16.4c^2\)
\(=\left(5a-3b\right)^2-16\left(a^2-b^2\right)\)
\(=25a^2-30ab+9b^2-16a^2+16b^2\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2\left(đpcm\right)\)
biến đổi vế trái
\(\Leftrightarrow\left(5a-3b\right)^2-\left(8c\right)^2\)
\(\Leftrightarrow25a^2-30ab+9b^2-64c^2\)
\(\Leftrightarrow25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(\Leftrightarrow\left(25a^2-16a^2\right)-30ab+\left(9b^2+16b^2\right)\)
\(\Leftrightarrow9a^2-30ab+25b^2\)
\(\Leftrightarrow\left(3a-5b\right)^2\) (điều cần c/m)
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