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a) 3x2 - 15x3 = 3x2.( 1 - 5x )
b) x.( 2x - 3 ) + y.( 3 - 2x ) = x.( 2x - 3 ) - y.( 2x - 3 ) = ( 2x - 3 ).( x - y )
c) ( 5x - y )2 - 4x2 = ( 5x - y )2 - ( 2x )2 = ( 5x - y - 2x ).( 5x - y + 2x ) = ( 3x - y ).( 7x - y )
d) x2 - 9y2 + 4 - 4x = ( x2 - 4x + 4 ) - 9y2 = ( x - 2 )2 - ( 3y )2 = ( x - 2 - 3y ).( x - 2 + 3y )
BÀI 1:
a) \(x^4+2x^2y+y^2=\left(x^2+y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)
\(=\left(3a+3b\right)\left(a-b\right)=3\left(a+b\right)\left(a-b\right)\)
c) \(\left(a^3-b^3\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left[a^2+ab+b^2+\left(a-b\right)\right]=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
d) \(\left(x^2+1\right)^2-4x^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x+1\right)^2\)
e) \(\left(y^3+8\right)+\left(y^2-4\right)=\left(y+2\right)\left(y^2-y+2\right)\)
f) \(1-\left(x^2-2xy+y^2\right)=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
g) \(x^4-1=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
h) ktra lại đề
m) \(\left(x-a\right)^4-\left(x+a\right)^4=-8ax\left(a^2+x^2\right)\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
Câu 1:
a) \(2x^2+5x-3=\left(2x^2+6x\right)-\left(x+3\right)\)
\(=2x\left(x+3\right)-\left(x+3\right)=\left(x+3\right)\left(2x-1\right)\)
b) \(x^4+2009x^2+2008x+2009\)
\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)
c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5-\sqrt{5}\\x=-5+\sqrt{5}\end{cases}}\)
Câu 1.
a) 2x2 + 5x - 3 = 2x2 + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )
b) x4 + 2009x2 + 2008x + 2009
= x4 + 2009x2 + 2009x - x + 2009
= ( x4 - x ) + ( 2009x2 + 2009x + 2009 )
= x( x3 - 1 ) + 2009( x2 + x + 1 )
= x( x - 1 )( x2 + x + 1 ) + 2009( x2 + x + 1 )
= ( x2 + x + 1 )[ x( x - 1 ) + 2009 ]
= ( x2 + x + 1 )( x2 - x + 2009 )
c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )
Câu 2.
3x2 + x - 6 - √2 = 0
<=> ( 3x2 - 6 ) + ( x - √2 ) = 0
<=> 3( x2 - 2 ) + ( x - √2 ) = 0
<=> 3( x - √2 )( x + √2 ) + ( x - √2 ) = 0
<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0
<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)
+) x - √2 = 0 => x = √2
+) 3( x + √2 ) + 1 = 0
<=> 3( x + √2 ) = -1
<=> x + √2 = -1/3
<=> x = -1/3 - √2
Vậy S = { √2 ; -1/3 - √2 }
Câu 3.
A = x( x + 1 )( x2 + x - 4 )
= ( x2 + x )( x2 + x - 4 )
Đặt t = x2 + x
A = t( t - 4 ) = t2 - 4t = ( t2 - 4t + 4 ) - 4 = ( t - 2 )2 - 4 ≥ -4 ∀ t
Dấu "=" xảy ra khi t = 2
=> x2 + x = 2
=> x2 + x - 2 = 0
=> x2 - x + 2x - 2 = 0
=> x( x - 1 ) + 2( x - 1 ) = 0
=> ( x - 1 )( x + 2 ) = 0
=> x = 1 hoặc x = -2
=> MinA = -4 <=> x = 1 hoặc x = -2
Sửa đề chút :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-x^3-y^3\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
c) x3 + y3 + z3 - 3xyz
= x3 + 3x2y + 3xy2 + y3 + z3 - 3xyz - 3x2y - 3xy2
= (x+y)3 + z3 - 3xy.( z+x+y)
= (x+y+z).[(x+y)2 - (x+y).z + z2 ] - 3xy.(x+y+z)
= (x+y+z). ( x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x+y+z) .(x2 + y2 + z2 - xy - xz -yz)
e) (a+b-c)2 - (a-c)2 - 2ab + 2bc
= (a+b-c - a+c).(a+b+c+a-c) - 2b.(a-c)
= b.(2a+b) - 2b.(a-c)
= b.(2a+b - a +c)
= b.( a+b+c)
xl bn nha! mk chỉ nghĩ đk 2 câu thoy, 1 câu bn kia làm r! 2 câu còn lại bn đợi người tiếp theo làm nhé
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow27x-2x-4x-27+2=0\)
\(\Leftrightarrow21x=25\)
\(\Leftrightarrow x=\frac{25}{21}\)
Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !
\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)
\(\Leftrightarrow-20x-12=56\)
\(\Leftrightarrow-20x=68\)
\(\Leftrightarrow x=-\frac{17}{5}\)
Tự check lại nhá
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
TL:
\(a,1-2m+m^2-x^2-4x-4\)
\(=\left(m-1\right)^2-\left(x-2\right)^2\)
\(=\left(m+x-3\right)\left(m-x+1\right)\)