Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)
b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)
\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x^3+1\right)}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{x-1}\cdot\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{-\left(x-1\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{-2\left(1+x+x^2\right)}{2x+3y}\)
a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)
\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)
\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)
b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4x^2-16.\dfrac{7x-2}{3x+6}\)
\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3x^3+3}{x-1}.\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x^3+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)
\(=\dfrac{2\left(2x+3y\right)\left(1-x\right)\left(1+x+x^2\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(2x+3y\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(x^2+x+1\right)}{2x+3y}\)
a) ( x2 - 25 )2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) ]2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]
= ( x - 5 )( x + 5 - 1 )( x - 5 )( x + 5 + 1 )
= ( x - 5 )2( x + 4 )( x + 6 )
b) ( 4x2 - 25 )2 - 9( 2x - 5 )2
= ( 4x2 - 25 )2 - 32( 2x - 5 )2
= ( 4x2 - 25 )2 - ( 6x - 15 )2
= [ ( 4x2 - 25 ) - ( 6x - 15 ) ][ ( 4x2 - 25 ) + ( 6x - 15 ) ]
= ( 4x2 - 25 - 6x + 15 )( 4x2 - 25 + 6x - 15 )
= ( 4x2 - 6x - 10 )( 4x2 + 6x - 40 )
= ( 4x2 + 4x - 10x - 10 )( 4x2 + 16x - 10x - 40 )
= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]
= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )
= ( 4x - 10 )2( x + 1 )( x + 4 )
c) 4( 2x - 3 )2 - 9( 4x2 - 9 )2
= 22( 2x - 3 )2 - 32( 4x2 - 9 )2
= ( 4x - 6 )2 - ( 12x2 - 27 )2
= [ ( 4x - 6 ) - ( 12x2 - 27 ) ][ ( 4x - 6 ) + ( 12x2 - 27 ) ]
= ( 4x - 6 - 12x2 + 27 )( 4x - 6 + 12x2 - 27 )
= ( -12x2 + 4x + 21 )( 12x2 + 4x - 33 )
= ( -12x2 + 18x - 14x + 21 )( 12x2 - 18x + 22x - 33 )
= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]
= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )
= ( x - 3/2 )2( -12x - 14 )( 12x + 22 )
d) a6 - a4 + 2a3 + 2a2
= a2( a4 - a2 + 2a + 2 )
= a2( a4 - 2a3 + 2a3 + 2a2 - 4a2 + a2 + 4a - 2a + 2 )
= a2[ ( a4 - 2a3 + 2a2 ) + ( 2a3 - 4a2 + 4a ) + ( a2 - 2a + 2 ) ]
= a2[ a2( a2 - 2a + 2 ) + 2a( a2 - 2a + 2 ) + 1( a2 - 2a + 2 ) ]
= a2( a2 + 2a + 1 )( a2 - 2a + 2 )
= a2( a + 1 )2( a2 - 2a + 2 )
e) ( 3x2 + 3x + 2 )2 - ( 3x2 + 3x - 2 )2
= [ ( 3x2 + 3x + 2 ) - ( 3x2 + 3x - 2 ) ][ ( 3x2 + 3x + 2 ) + ( 3x2 + 3x - 2 ) ]
= ( 3x2 + 3x + 2 - 3x2 - 3x + 2 )( 3x2 + 3x + 2 + 3x2 + 3x - 2 )
= 4( 6x2 + 6x )
= 4.6x( x + 1 )
= 24( x + 1 )
c)\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
d)\(\dfrac{a^2}{4}-2a+4=\left(\dfrac{a}{2}-2\right)^2\)
e) \(4y^2-9x^2=\left(2y-3x\right)\left(2y+3x\right)\)
f)\(9y^2-\dfrac{1}{4}=\left(3y-\dfrac{1}{2}\right)\left(3y+\dfrac{1}{2}\right)\)
g)\(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4xa+4a^2\right)\)
bn chép lại đề nhé
a/ \(=\left(x+y\right)^2-4x^2y^2=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
b/ \(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[-\left(b+c\right)^2+a^2\right]\)
\(=\left(b+c-a\right)\left(b+c+a\right)^2\left(a-b-c\right)\)
c/ \(=2a^2+2b^2-2c^2+4ab=2\left[\left(a^2+b^2+2ab\right)-c^2\right]\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
d/ \(=\left(4x^2-25\right)^2-9\left(4x^2-20x+25\right)\)
\(=\left(4x^2-25\right)^2-9\left(4x^2+25\right)+180x\)
tới đây bạn đặt a= 4x^2 -25 rồi làm típ nha, mình lười quá ><
e/ tương tự câu d nha bạn
f/ \(=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)
\(=a^4\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)\)
\(=a^2\left(a+1\right)\left(a^2+2\right)\)
g/ đặt \(a=3x^2+3x+2\) khi đó biểu thức trở thành
\(a^2-\left(a+4\right)^2=a^2-a^2-8a-16\)
\(=-8a-16=-8\left(3x^2+3x+2-8\right)=-8\left(3x^2+3x-6\right)\)
\(=-24\left(x^2+x-2\right)=-24\left(x-1\right)\left(x+2\right)\)
xong rùi nha bn. Chúc bn hc tốt (xin lỗi tại có mấy câu mình lười nha)
\(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2\)
\(=\left(x-2xy+y\right)\left(x+2xy+y\right)\)
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
1. x^3-19x-30
=x^3-25x+6x-30
=x(x^2-25)+6(x-5)
=x(x+5)(x-5)+6(x-5)
=(x-5)(x^2+5x+6)
=(x-5)(x^2+2x+3x+6)
=(x-5)[x(x+2)+3(x+2)]
=(x-5)(x+2)(x+3)
2.
a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) ------------(1)
CẦn chứng minh:
2(a^4 + b^4 + c^4) = (a² + b² + c²)²
<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²)
<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²)
<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) )
<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1))
<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²)
<=> 8.(ab²c + bc²a + a²bc) = 0
<=> 8abc.(a + b + c) = 0
<=> 0 = 0 (đúng), Vì a + b + c = 0
=> Đpcm
bạn áp dụng các hằng đẳng thức là Ok:
a) \(-4x^2+12xy-9y^2+25=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+25\)
= \(-\left(2x-3y\right)^2+25=\left(5-2x+3y\right)\left(5+2x-3y\right)\)
b) \(x^3-3x^2+3x-1=\left(x^3-1\right)-\left(3x^2-3x\right)=\left(x-1\right)\left(x^2+x+1^2\right)-3x\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x+1^2-3x\right)=\left(x-1\right)\left(x^2-2x+1\right)=\left(x-1\right)\left(x-1\right)^2=\left(x-1\right)^3\)
c) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a^2-1\right)+2\left(a+1\right)\right]\)
= \(a^2\left[a^2\left(a+1\right)\left(a-1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3+a^2+2\right)\)
Thanks bạn hiền