Rút gọn biểu thức :
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
b) \(\left(4x^2-3y\right).2y-\left(3x^2-4y\right).3y\)
c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
Rút gọn các biểu thức sau:
a/ \(\left(3x-1\right)^2-2\left(2-5x\right)-2\left(x^2^{^{ }}+x-1\right)\left(x-\dfrac{1}{2}\right)\)
b/\(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
c/\(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
d/\(\left(3a-1\right)^2+2\left(9a^2-1\right)\left(3a+1\right)\)
e/\(\left(3x-4\right)^2+\left(4-x\right)^2-2\left(3x-4\right)\left(x-4\right)\)
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b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Rút gọn biểu thức :
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
b) \(\left(4x^2-3y\right).2y-\left(3x^2-4y\right).3y\)
c) \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
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a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+16\)
\(=-11x+16\)
b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(=8x^2y-6y^2-\left(9x^2y-12y^2\right)\)
\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)
c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2.3y-y+y^2+y^3+y\)
\(=9y^3+y^2+y^3=10y^3+y^2\)
Chúc bạn học tốt!!!
a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+16\)
\(=-11x+16\)
b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=-x^2y+6y^2\)
c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2.3y-y\left(1-y-y^2-1\right)\)
\(=9y^3-y\left(-y-y^2\right)\)
\(=9y^3+y^2+y^3=10y^3+y^2\)
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+16\)
\(=3x^2+5x^2-8x^2-6x-5x+16\)
\(=-11x+16\)
b) \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right).3y\)
\(=9x^2y-6y^2-\left(9x^2y-12y^2\right)\)
\(=9x^2y-6y^2-9x^2y+12y^2\)
\(=9x^2y-9x^2y-6y^2+12y^2\)
\(=6y^2\)
c) \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2\left(2y-1\right)+3y^3+3y^2-y+y^2+y^3+y\)
\(=6y^3-3y^2+3y^3+3y^2-y+y^2+y^3+y\)
\(=6y^3+3y^3+y^3-3y^2+3y^2+y^2-y+y\)
\(=10y^3+y^2\)
Bài 1 :Tìm x,y ,biết :
a) \(\left(3x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=2014\)
b) \(5x^2+4xy+4y^2+4x+1=0\)
Bài 2 : Chứng minh rằng các biểu thức sau không phụ thuộc vào các biến x,y:
D = \(\left(2x-3y\right)^2-\left(3y-2\right)\left(3y+2\right)-\left(1-2x\right)^2+4x\left(3y-1\right)\)
Bài 1 :
a) \(\left(3x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=2014\)
\(\Leftrightarrow9x^2-6x+1-\left(9x^2-4\right)=2014\)
\(\Leftrightarrow-6x=2009\)
\(\Leftrightarrow x=-\dfrac{2009}{6}=-334\dfrac{5}{6}\)
b) \(5x^2+4xy+4y^2+4x+1=0\)
\(\Leftrightarrow\left(x^2+4xy+4y^2\right)+\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow\left(x+2y\right)^2+\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{4}\end{matrix}\right.\)
Bài 2 :
Ta có :
\(D=\left(4x^2-12xy+9y^2\right)-\left(9y^2-4\right)-\left(1-4x+4x^2\right)+12xy-4x\)
\(=4x^2-12xy+9y^2-9y^2+4-1+4x-4x^2+12xy-4x=3\)
Vậy biểu thức D không phụ thuộc vào các biến x,y
Bài 1 :Rút gọn
\(\left(4x^2-3y\right)a2y-\left(3x^2-4y\right)3y\)
\(4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(2ax^2-a\left(1+2x^2\right)-\left\{a-x\left(x+a\right)\right\}\)
Bài 2:Tìm x
a)\(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+1=0\)
b)\(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
Bài 3:Rút gọn
\(x\left(1+x+x^2+...+x^9\right)-\left(1+x+x^2+...+x^9\right)\)
Thực hiện phép tính:
a) \(\dfrac{2}{5}xy\left(x^2y-5x+10y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x+y\right)\)
c) \(\left(x+3y\right)^2\)
d) \(\left(4x-y\right)^3\)
e) \(\left(x^2-2y\right)\left(x^2+2y\right)\)
g) \(18x^4y^2z:10x^4y\)
h) \(\left(x^3y^3+\dfrac{1}{2}x^2y^3-x^3y^2\right):\dfrac{1}{3}x^2y^2\)
i) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
k) \(\dfrac{5x-1}{3x^2y}+\dfrac{x+1}{3x^2y}\)
l) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
m) \(\dfrac{2x+3}{10x-4}+\dfrac{5-3x}{4-10x}\)
n) \(\dfrac{x}{x^2+2x+1}+\dfrac{3}{5x^2-5}\)
o) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
p) \(\dfrac{4x+2}{15x^3y}\dfrac{5y-3}{9x^2y}+\dfrac{x+1}{5xy^3}\)
q) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
r) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
x) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\)
y) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
z) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
t) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
w) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c: \(=x^2+6xy+9y^2\)
e: \(=x^4-4y^2\)
Bài 4.Rút gọn các biểu thức sau
1) \(4x^2\left(5x^3-2x+3\right)\)
2) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
3) \(\left(5x^2-4x\right)\left(x-2\right)\)
4) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
1) 4x2( 5x3 - 2x + 3 ) = 20x5 - 8x3 + 12x2
2) 3y2( 4y3 + 2/3y2 - 1/3 ) = 12y5 + 2y4 - y2
3) ( 5x2 - 4x )( x - 2 ) = 5x3 - 14x2 + 8x
4) ( 3x - 5 )( 2x + 11 ) - ( 2x + 3 )( 3x + 7 )
= 6x2 + 23x - 55 - ( 6x2 + 23x + 21 )
= 6x2 + 23x - 55 - 6x2 - 23x - 21
= -76
Rút gọn biểu thức rồi tính giá trị:
a) \(\frac{x^2y\left(y-x\right)+xy^2\left(x-y\right)}{3y^2-3x^2}\) ,với x = -3 ; y =\(\frac{1}{2}\)
b) \(\frac{\left(8x^3-y^3\right)\left(4x^2-y^2\right)}{\left(2x+y\right)\left(4x^2-4xy+y^2\right)}\)với x = 2; y =\(\frac{-1}{2}\)