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\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)
\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
\(2)3x+4-x^2-4x\)
\(=3(x+4)-\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(3)5x^2-2y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)[5(x+y)-10]\)
Còn lại bn lm nốt nha!
= ( 2x - 5 )(2x + 5) - ( 2x- 5 )(2x + 7 )
= (2x - 5 ) [ 2x+ 5 - ( 2x+ 7 ) ]
= ( 2x- 5 ) ( 2x + 5 - 2x - 7 )
= - 2( 2x- 5 )
7, 4x mũ 2 - 12x + 9 - y mũ 2 = -(y-2x+3) (y+2x-3)
8, 16x mũ 2 - 4y mũ 2 + 4y - 1 = -(2y - 4x - 1) (2y+4x-1)
9, 25 - x mũ 2 - 12x - 36 = -(x+1) (x+11)
10, x mũ 2 - 9 - 5 ( x + 3 ) = (x-8) (x+3)
bạn k cho mình nha
chúc bạn học tốt :))))
bạn kham khảo link, mình đã làm rồi nhé
Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM
\(a,3x^2+2x=x\left(3x+2\right)\)
\(b,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)
\(c,4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(d,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
\(e,x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
a ) 3x2 + 2x
= x. ( 3x + 2 )
b ) 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x.( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
c ) 4x2 - 25
= ( 2x + 5 ) ( 2x - 5 )
d ) x2 + 6x + 5
= x2 + x + 5x + 5
= x.( x + 1 ) + 5.( x + 1 )
= ( x + 1 ) ( x + 5 )
e ) x2 - y2 + 2y - 1
= x2 - ( y - 1 )2
= ( x - y + 1 ) ( x + y - 1 )
f ) x3 - 3x + 2
= x3 + 2x2 - 2x2 - 4x + x + 2
= x2 ( x + 2 ) - 2x ( x + 2 ) + ( x + 2 )
= ( x + 2 ) ( x2 - 2x + 1 )
= ( x + 2 ) ( x - 1 )2
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
a) \(2x^2-2y^2=2\left(x^2-y^2\right)=2\left(x-y\right)\left(x+y\right)\)
b) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\)
c) \(x^2-4x=x\left(x-4\right)\)
d) \(x^2+10x+25=x^2+2.5x+5^2=\left(x+5\right)^2\)
e) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y+3\right)\left(x-y-3\right)\)
\(2x^2-2y^2=2.\left(x^2-y^2\right)=2.\left(x-y\right)\left(x+y\right)\)
\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(x^2-4x=x.\left(x-4\right)\)
\(x^2+10x+25=x^2+2.x.5+5^2=\left(x+5\right)^2\)
\(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
Tham khảo nhé~
1/\(\left(x^2-25\right)^2-\left(x-5\right)^2\)
<=>\(\left[\left(x-5\right)\left(x+5\right)\right]^2-\left(x-5\right)^2\)
<=>\(\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]\)
2/\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
<=>\(\left[\left(2x-5\right)\left(2x+5\right)\right]^2-9\left(2x-5\right)^2\)
<=>\(\left(2x-5\right)\left[\left(2x+5\right)^2-9\right]\)
#hoctot<3#