cho tam giác abc có trung tuyến am , i là một điểm thuộc đoạn thẳng am bi cắt ac ở d a, nếu ad=1/2dc khi đó hãy chứng minh i là trung điểm của am b, nếu i là trung điểm của am khi đó cm ad = 1/2 dc id = 1/4 bd c, nếu ad = 1/2 dc lhi đó trên cạnh ab ấy điểm e sao cho ab =3ae chứng minh bd, ce, am đồng quy
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y^2+y+y+1\)
\(=y^2+2y+1\)
\(=\left(y+1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+3\right)^2-\left(2x+6\right)\left(1-3x\right)+\left(3x-1\right)^2\)
\(=\left(x+3\right)^2-2\left(x+3\right)\left(1-3x\right)+\left(1-3x\right)^2\)
\(=\left(x+3-1+3x\right)^2=\left(4x+2\right)^2=16x^2+16x+4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(x^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
2, \(x^3-7x+6=x^3-x^2+x^2-x-6x+6\)
\(=x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-1\right)=\left(x^2+x-6\right)\left(x-1\right)=\left(x-2\right)\left(x+3\right)\left(x-1\right)\)
3, \(x^2+4xy+3y^2=x^2+4xy+4y^2-y^2\)
\(=\left(x+2y\right)^2-y^2=\left(x+2y-y\right)\left(x+2y+y\right)=\left(x+y\right)\left(x+3y\right)\)
4, \(x^2-8x+12=x^2-2x-6x+12\)
\(=x\left(x-2\right)-6\left(x-2\right)=\left(x-6\right)\left(x-2\right)\)
Trả lời:
1, x4 - 5x2 + 4
= x4 - 4x2 - x2 + 4
= ( x4 - 4x2 ) - ( x2 - 4 )
= x2 ( x2 - 4 ) - ( x2 - 4 )
= ( x2 - 4 )( x2 - 1 )
= ( x - 2 )( x + 2 )( x - 1 )( x + 1 )
2, x3 - 7x + 6
= x3 - 7x + 6 + x2 - x2
= x3 - 6x - x + 6 + x2 - x2
= ( x3 - x2 ) + ( x2 - x ) - ( 6x - 6 )
= x2 ( x - 1 ) + x ( x - 1 ) - 6 ( x - 1 )
= ( x - 1 )( x2 + x - 6 )
= ( x - 1 )( x2 - 2x + 3x - 6 )
= ( x - 1 )[ x ( x - 2 ) + 3 ( x - 2 ) ]
= ( x - 1 )( x + 3 )( x - 2 )
3, x2 + 4xy + 3y2
= x2 + 4xy + 4y2 - y2
= ( x2 + 4xy + 4y2 ) - y2
= ( x + 2y )2 - y2
= ( x + 2y - y )( x + 2y + y )
= ( x + y )( x + 3y )
4, x2 - 8x + 12
= x2 - 6x - 2x + 12
= x ( x - 6 ) - 2 ( x - 6 )
= ( x - 6 )( x - 2 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
x4 - 3x3 + 3x2 - x
= x ( x3 - 3x2 + 3x - 1 )
= x ( x - 1 )3
Ta có :
\(x^4-3x^3+3x^2-x\)
\(=x\left(x^3-3x^2+3x-1\right)\)
\(=x\left(x-1\right)^3\)
Vậy ..........
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=x^2+y^2=\left(x+y\right)^2-2xy\)
\(\Rightarrow A=a^2-2b\)
b, \(B=x^2-y^2=\left(x+y\right)^2-4xy\)
\(\Rightarrow A=a^2-4b\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(35x^2-79x-12=35x^2-84x+5x-12\)
\(=5x\left(7x+1\right)-12\left(7x+1\right)=\left(5x-12\right)\left(7x+1\right)\)
2, \(20x^2+45x-24x-54=5x\left(4x+9\right)-6\left(4x+9\right)\)
\(=\left(5x-6\right)\left(4x+9\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)
\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)
\(=\left(4x^2-6x-5\right)\left(4x^2+6x-40\right)=2\left(x+4\right)\left(2x-5\right)\left(4x^2-6x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
8x3 - 24x2 + 18x
= 2x ( 4x2 - 12x + 9 )
= 2x [ (2x)2 - 2.2x.3 + 32 ]
= 2x ( 2x - 3 )2