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a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
Bài 4:
1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)
=>\(x^3-1-x^3-6x=11\)
=>-6x-1=11
=>-6x=11+1=12
=>\(x=\dfrac{12}{-6}=-2\)
2: \(16x^2-\left(3x-4\right)^2=0\)
=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)
=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)
=>(x+4)(7x-4)=0
=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)
3: \(x^3-x^2-3x+3=0\)
=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)
=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))
=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)
=>\(x^2+4x+4=x^2-1\)
=>4x+4=-1
=>4x=-5
=>\(x=-\dfrac{5}{4}\left(nhận\right)\)
5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)
=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)
=>3x+1=0
=>3x=-1
=>\(x=-\dfrac{1}{3}\left(nhận\right)\)
6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)
=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-x-3}{x}=1\)
=>-x-3=x
=>-2x=3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)
\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)