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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
\(\Rightarrow\left(đpcm\right)\)
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\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)
\(\left(5a-3b\right)^2-\left(3a-5b\right)^2-64c^2=0\)
\(\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)-64c^2=0\)
\(\left(2a+2b\right)\left(8a-8b\right)-64c^2=0\)
\(16a^2-16ab+16ab-16b^2-64c^2=0\)
\(16a^2-16b^2-64c^2=0\)
\(16\left(a^2-b^2\right)-64c^2=0\)
\(16\times4c^2-64c^2=0\)
\(64c^2-64c^2=0\left(dpcm\right)\)
\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left(5a-3b\right)^2-\left(8c\right)^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2\)
Bài 1.
a) 2x - x2
= x(2 - x)
b) 16x2 - y2
= (4x + y)(4x - y)
c) xy + y2 - x - y
= (xy + y2) - (x + y)
= y(x + y) - (x + y)
= (y - 1)(x + y)
d) x2 - x - 12
= x2 + 3x - 4x - 12
= (x2 + 3x) - (4x + 12)
= x(x + 3) - 4(x + 3)
= (x - 4)(x + 3)
Bài 2.
(2x + 3y)(2x - 3y) - (2x - 1)2 + (3y - 1)2
= (2x + 3y)(2x - 3y) + [(3y - 1)2 - (2x - 1)2]
= (2x + 3y)(2x - 3y) + (3y - 1 + 2x - 1)(3y - 1 - 2x + 1)
= (2x + 3y)(2x - 3y) + (3y + 2x - 2)(3y - 2x)
= (2x + 3y)(2x - 3y) - (2x + 3y - 2)(2x - 3y)
= (2x - 3y)(2x + 3y - 2x - 3y + 2)
= 2.(2x + 3y)
Thay x = 1; y = -1 và biểu thức đại số, ta có:
2[2.1 + 3.(-1)]
= 2(2 - 3)
= 2.(-1) = -2
Bài 3
a) 9x2 - 3x = 0
3x(3x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}3x=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\3x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) x2 - 25 - (x + 5) = 0
(x2 - 25) - (x + 5) = 0
(x - 5)(x + 5) - (x + 5) = 0
(x - 5 - 1)(x + 5) = 0
(x - 6)(x + 5) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
c) x2 + 4x + 3 = 0
x2 + x + 3x + 3 = 0
(x2 + x) + (3x + 3) = 0
x(x + 1) + 3(x + 1) = 0
(x + 3)(x + 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d) (3x - 1)(2x - 7) - (x + 1)(6x - 5) = 16
6x2 - 21x - 2x + 7 - 6x2 + 5x - 6x + 5 - 16 = 0
-24x - 4 = 0
\(\Rightarrow\)-24x = 4
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Bài 1:Phân tích đa thức thành nhân tử
a,2x−x2
=x(2-x)
b,
16x2−y2
=(4x-y)(4x+y)
c,xy+y2−x−y
=(xy+y2)-(x+y)
=y(x+y)-(x+y)
=(x+y)(y-1)
d,
x2−x−12
=x2-4x+3x-12
=(x2-4x)+(3x-12)
=x(x-4)+3(x-4)
=(x-4)(x+3)
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)=\left(3a-5b\right)^2\\ 25a^2-15ab-20ac-15ab+9b^2+12bc+20ac-12bc-16c^2=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2-30ab=9a^2-30ab+25b^2\\ \Leftrightarrow25a^2+9b^2-16c^2=9a^2+25b^2\\ \Leftrightarrow25a^2-9a^2=-9b^2+25b^2+16c^2\\ \Leftrightarrow16a^2-=16b^2+16c^2\\ \Leftrightarrow a^2=b^2+c^2\)
Vậy ...
\(a^2-b^2-c^2=0\Rightarrow c^2=a^2-b^2\)
\(\left(5a-3b+4c\right)\left(5a-3b-4c\right)\)
\(=\left(5a-3b\right)^2-\left(4c\right)^2\)
\(=25a^2-30ab+9b^2-16c^2\)
\(=25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a\right)^2-2.3a.5b+\left(5b\right)^2=\left(3a-5b\right)^2\)
Chúc bạn học tốt.
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 )
\(\Rightarrow\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)