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\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)=\left(3a-5b\right)^2\\ 25a^2-15ab-20ac-15ab+9b^2+12bc+20ac-12bc-16c^2=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2-30ab=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2=9a^2+25b^2\\ \Leftrightarrow25a^2-9a^2=-9b^2+25b^2+16c^2\\ \Leftrightarrow16a^2-=16b^2+16c^2\\ \Leftrightarrow a^2=b^2+c^2\)
Vậy ...
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
\(\Rightarrow\left(đpcm\right)\)
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
xét hiệu\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-\left(3a-5b\right)^2-64c^2=0\)
\(\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)-64c^2=0\)
\(\left(2a+2b\right)\left(8a-8b\right)-64c^2=0\)
\(16a^2-16ab+16ab-16b^2-64c^2=0\)
\(16a^2-16b^2-64c^2=0\)
\(16\left(a^2-b^2\right)-64c^2=0\)
\(16\times4c^2-64c^2=0\)
\(64c^2-64c^2=0\left(dpcm\right)\)
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2\)
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)=\left(3a-5b\right)^2\)
\(\Leftrightarrow25a^2-15ab-20ac-15ab+9b^2+12bc+20ac-12bc-16c^2=9a^2-30ab+25b^2\)
\(\Leftrightarrow25a^2-30ab+9b^2-16c^2=9a^2-30ab+25b^2\)
\(\Leftrightarrow25a^2+9b^2-16c^2=9a^2+25b^2\)
\(\Leftrightarrow16a^2=16c^2+16b^2\)
\(\Rightarrow a^2=b^2+c^2\)
\(\Rightarrow\Delta\) với 3 cạnh a, b, c vuông
\(\Rightarrow\Delta\) có độ dài 3 cạnh trên là \(\Delta\) vuông ( đpcm )
Vậy...
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 )
\(\Rightarrow\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)
Bài 1:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
\(\Rightarrowđpcm\)
Bài 2:
Ta có: \(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=25a^2-30ab+9b^2-64c^2\)
\(=25a^2-30ab+9b^2-16a^2+16b^2\left(a^2-b^2=4c^2\right)\)
\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2=VP\)
\(\Rightarrowđpcm\)
Bài 1:
\(B=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left(0,625-0,5+\frac{5}{11}+\frac{5}{12}\right)}+\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}\)
\(=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-\left[5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)\right]}+\frac{3}{5}\)
\(=\frac{-3}{5}+\frac{3}{5}\)
\(=0\)
Bài 2:
b) Giải:
Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{c^6}{d^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (1)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{a+c}{b+d}\right)^6=\frac{a^6}{b^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\left(đpcm\right)\)
\(a^2-b^2-c^2=0\Rightarrow c^2=a^2-b^2\)
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)\)
\(=\left(5a-3b\right)^2-\left(4c\right)^2\)
\(=25a^2-30ab+9b^2-16c^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a\right)^2-2.3a.5b+\left(5b\right)^2=\left(3a-5b\right)^2\)
Chúc bạn học tốt.