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\(b,x^2+4x+3=x^2+3x+x+3.\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(c,16x-5x^2-3=x-5x^2+15x-3\)
\(=x\left(1-5x\right)+3\left(5x-1\right)\)
\(=\left(x+3\right)\left(1-5x\right)\)
\(d,x^4+4=x^4+4x^2+4-4x^2=\left(x+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
Câu a trước đi ạ ^^
a) 7x - 6x2 - 2
= - 6x2 + 7x - 2
= (- 6x2 + 3x) + (4x - 2)
= 3x (- 2x + 1) + 2 (2x-1)
= - 3x ( 2x -1) + 2 (2x - 1)
= ( 2x -1 ) ( - 3x +2 )
B1 :
a, B = (x+1)^2+(y-2)^2 = (99+1)^2+(102-2)^2 = 100^2+100^2 = 20000
b, = (2x^2+16x+32)-2y^2
= 2.(x+4)^2-2y^2
= 2.[(x+4)^2-y^2] = 2.(x+4-y).(x+4+y)
c, <=> (x^2-3x)+(2x-6) = 0
<=> (x-3).(x+2) = 0
<=> x-3=0 hoặc x+2=0
<=> x=3 hoặc x=-2
B2 :
P = (3-x).(x+3)/x.(x-3) = -(x+3)/x = -x-3/x
k mk nha
Bai 1
a)B=(x+1)2+(y-2)2
Voi x=99,y=102
=>B= 1002+1002
=20000
b)\(2x^2-2y^2+16x+32\)
=\(2\left[\left(x^2+8x+16\right)-y^2\right]\)
=\(2\left[\left(x+4\right)^2-y^2\right]\)
=2(x-y+4)(x+y+4)
c)\(x^2-3x+2x-6=0\)
=>x(x-3)+2(x-3)=0
=>(x-3)(x+2)=0
=>x=-2;3
Bai 2
\(P=\frac{9-x^2}{x^2-3x}\)
=\(-\frac{x^2-9}{x\left(x-3\right)}\)
=\(-\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
=\(\frac{-x-3}{x}\)
a) 16x2 - 9
= ( 4x )2 - 32
= ( 4x - 3 )( 4x + 3 )
b) 9a2 - 25b4
= ( 3a )2 - ( 5b2 )2
= ( 3a - 5b2 )( 3a + 5b2 )
c) 81 - y4
= 92 - ( y2 )2
= ( 9 - y2 )( 9 + y2 )
= ( 32 - y2 )( 9 + y2 )
= ( 3 - y )( 3 + y )( 9 + y2 )
d) ( 2x + y )2 - 1
= ( 2x + y )2 - 12
= ( 2x + y - 1 )( 2x + y + 1 )
e) ( x + y + z )2 - ( x - y - z )2
= [ x + y + z - ( x - y - z ) ][ x + y + z + ( x - y - z ) ]
= [ x + y + z - x + y + z ][ x + y + z + x - y - z ]
= [ 2y + 2z ].2x
= 2[ y + z ].2x
= 4x[ y + z ]
bài 1
a)\(x^2+5x+6=\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)
a: Ta có: \(a^5-ax^4+a^4x-x^5\)
\(=a\left(a^4-x^4\right)+x\left(a^4-x^4\right)\)
\(=\left(a-x\right)\left(a+x\right)\left(a^2+x^2\right)\cdot\left(a+x\right)\)
\(=\left(a-x\right)\cdot\left(a+x\right)^2\cdot\left(a^2+x^2\right)\)