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1/ 2a + 2b = 2( a + b )
2/ 3a - 6b - 9c = 3( a - 2b - 3c )
3/ 5ax - 15ay + 20a = 5a( x - 3y + 4 )
4/ 3a2x - 6a2y + 12a = 3a( ax - 2ay + 4 )
5/ 4a( x - 5 ) - 2( 5 - x ) = 4a( x - 5 ) + 2( x - 5 ) = ( x - 5 )( 4a + 2 ) = ( x - 5 )2( 2a + 1 )
6. -3a( x - 3 ) + ( 3 - x ) = 3a( 3 - x ) + 1( 3 - x ) = ( 3a + 1 )( 3 - x )
7/ xm+1 - xm = xm( x + 1 )
8/ xm+2 - x2 = x2( xm - 1 )
\(1.4a\left(x-5\right)-2\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(4a+2\right)\)
\(\Leftrightarrow2\left(x-5\right)\left(2a+1\right)\)
\(2.-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(\Leftrightarrow\left(x-3\right)\left(a^2-3a\right)\)
\(\Leftrightarrow a\left(x-3\right)\left(a-3\right)\)
\(3.\text{ }\text{ }\text{ }2a^2b.\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(2a^2b+4a^3b\right)\)
\(\Leftrightarrow2a^2b\left(x+y\right)\left(1+2a\right)\)
1, 4a(x-5) +2(x-5) 2, 3a(3-x)-a^2(3-x) 3, 2a^2 b(x+y)+4a^3 b(x+y)
(4a+2)(x-5) (3-a)a(3-x) 2a^2 b (1+2a)(x+y)
minh khong lam het duoc vi khong co thoi gian
1 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
2 ) \(a^2\left(x+y\right)-b^2\left(x+y\right)\)
\(=\left(a^2-b^2\right)\left(x+y\right)\)
\(=\left[\left(a-b\right).\left(a+3\right)\right]\left(x+y\right)\)
3 ) \(6a^2-3a+12ab\)
\(=3a.2a-3a+3a.4b\)
\(=3a.\left(2a-1+4b\right)\)
4 ) \(2x^2y^4-2x^4y^2+6x^3y^3\)
\(=2x^2y^2.y^2-2x^2y^2.x^2+2x^2y^2.3xy\)
\(=2x^2y^2\left(y^2-x^2+3xy\right)\)
5 ) \(\left(x+y\right)^3-x\left(x+y\right)^2\)
\(=\left(x+y\right)^2.\left(x+y-x\right)\)
\(=\left(x+y\right)^2.y\)
1)a(m+n)+b(m+n)
=(a+b)(m+n)
2)a2(x+y)-b2(x+y)
=(a2-b2)(x+y)
3)6a2-3a+12ab
=3a.2a-3a.(1-4b)
=3a.(2a-1+4b)
5)(x+y)3-x(x+y)2
=(x+y)(x+y)2-x(x+y)2
=(x+y)2(x+y-x)
câu a nè = (4x-1)(2x-3)
câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx)
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
Bài 2:
a: =>6x^2-4x-10=0
=>3x^2-2x-5=0
=>3x^2-5x+3x-5=0
=>(3x-5)(x+1)=0
=>x=-1 hoặc x=5/3
b: =>(x+1)(x+2)=0
=>x=-1 hoặc x=-2
Bài 3:
\(=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}\)
Dấu = xảy ra khi x=-1/2
a: \(=\dfrac{5}{2}x-2x+\dfrac{7}{2}=\dfrac{1}{2}x+\dfrac{7}{2}\)
b: \(=\dfrac{-1}{4}x^4-3x^2+\dfrac{9}{4}x\)
c: \(=\dfrac{1}{5}x+\dfrac{1}{15}xy+\dfrac{7}{10}x^2\)
d: \(=-9x^3-1-12y+27xy\)
Bác google được sinh ra để làm gì, đăng nhiều vc, google có hết mà ;v
Bài 1,2,3,4 đơn giản, tự làm :v
7) \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{1}{3abc}=\dfrac{1}{3}\)
P/S: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
5) ĐK: a>b>0
\(3a^2+3b^2=10ab\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)
Tự phân tích
Mà a>b>0=> Chọn a=3b
Thay vào
Bài 6 tương tự bài 5
Có bất mãn chỗ nào thì ib nha bạn :))
1) \(4a\left(x-5\right)-2\left(5-x\right)\)
\(=4a\left(x-5\right)+2\left(x-5\right)\)
\(=2\left(x-5\right)\left(2a+1\right)\)
2) \(-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(=-3a\left(x-3\right)+a^2\left(x-3\right)\)
\(=a\left(x-3\right)\left(-3+a\right)\)
3) \(2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(=2a^2b\left(x+y\right)+4a^3b\left(x+y\right)\)
\(=2a^2b\left(x+y\right)\left(1+2a\right)\)
4) \(-3a\left(x-3\right)-a^2\left(3-a\right)\)
Mình nghĩ câu này đề sai và hình như nó là câu 2 thì phải
5) \(x^{m+1}-x^m\)
\(=x^m.x-x^m\)
\(=x^m\left(x-1\right)\)
6) \(x^{m+1}+x^m\)
\(=x^m.x+x^m\)
\(=x^m\left(x+1\right)\)
7) \(x^{m+2}-x^m\)
\(=x^m.x^2-x^m\)
\(=x^m\left(x^2-1\right)\)
\(=x^m\left(x-1\right)\left(x+1\right)\)
8) \(x^{m+2}-x^2\)
\(=x^m.x^2-x^2\)
\(=x^2\left(x^m-1\right)\)
9) \(x^{m+2}-x^{m+1}\)
\(=x^{m+1}.x-x^{m+1}\)
\(=x^{m+1}\left(x-1\right)\)