Viết các biểu thức sau thành đa thức:
a) \(\left( {3x - 5} \right)\left( {3x + 5} \right)\) b) \(\left( {x - 2y} \right)\left( {x + 2y} \right)\) c) \(\left( { - x - \dfrac{1}{2}y} \right)\left( { - x + \dfrac{1}{2}y} \right)\)
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a) \(\left(x-5\right)\left(a^2+5a+25\right)\)
\(=a^3-5^3\)
\(=a^3-125\)
b) \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(=x^3+\left(2y\right)^3\)
\(=x^3+8y^3\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
a)
\(\begin{array}{l}{\left( {x - 2y} \right)^3} + {\left( {x + 2y} \right)^3}\\ = {x^3} - 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} - {\left( {2y} \right)^3} + {x^3} + 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} + {\left( {2y} \right)^3}\\ = 2{x^3} + 24x{y^2}\end{array}\)
b)
\(\begin{array}{l}{\left( {3x + 2y} \right)^3} + {\left( {3x - 2y} \right)^3}\\ = {\left( {3x} \right)^3} + 3.{\left( {3x} \right)^2}.2y + 3.3x{\left( {2y} \right)^2} + {\left( {2y} \right)^3} + {\left( {3x} \right)^3} - 3.{\left( {3x} \right)^2}.2y + 3.3x{\left( {2y} \right)^2} - {\left( {2y} \right)^3}\\ = 54{x^3} + 72x{y^2}\end{array}\)
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
a\(=3x^2-6x+6x-3x^2+5=5\)=>ko phụ thuộc vào biến x
b,\(=2x^2y-2xy^2+2xy^2-x^2y-x^2y=0\)=>ko phụ thuộc vào biến ,x,y
a: \(=4x^2-25-4x^2+12x-9-12x=-34\)
b: \(=8y^3-12y^2+6y-1-2y\left(4y^2-12y+9\right)-12y^2+12y\)
\(=8y^3-24y^2+18y-1-8y^3+24y^2-18y=-1\)
c: \(=x^3+27-x^3-20=7\)
d: \(=3y\left(9y^2+12y+4\right)-27y^3+1-36y^2-12y-1\)
\(=27y^3+36y^2+12y-27y^3-36y^2-12y\)
=0
a) \(5\left(x+2y\right)-15x\left(x+2y\right)=\left(x+2y\right)\left(5-15x\right)\\ =5\left(x+2y\right)\left(1-3x\right)\)
b) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2\\=\left(2x-3\right)^2\)
c) \(\left(3x-2\right)^3-3\left(x-4\right)\left(x+4\right)+\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =27x^3-54x^2+18x-8-3\left(x^2-16\right)+x^3-9x^2+27x-27-\left(x^3+1\right)\\=27x^3-54x^2+18x-8-3x^2+48+x^3-9x^2+27x-27-x^3-1\\ =27x^3-57x^2+36x+12\\ =3\left(3x^3-19x^2+12x+4\right)\)
Câu 1: \(3x+2\left(5-x\right)=0\)
\(\Rightarrow3x+10-2x=0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\).
Câu 2: \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=3\)
\(\Rightarrow2x\left(5-3x\right)-2x\left(5-3x\right)-3\left(x-7\right)=0\)
\(\Rightarrow\left(2x-2x\right)\left(5-3x\right)-3\left(x-7\right)=3\)
\(\Rightarrow-3\left(x-7\right)=3\)
\(\Rightarrow x-7=-1\)
\(\Rightarrow x=6.\)
Câu 3:
Áp dụng hằng đẳng thức mở rộng có:
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+b^3+c^3-3abc.\)
Câu 4: \(3x^2\left(3x^2-2y^2\right)-\left(3x^2-2y^2\right)\left(3x^2+2y^2\right)\)
\(=\left(3x^2-2y^2\right)\left[3x^2-\left(3x^2+2y^2\right)\right]\)
\(=\left(3x^2-2y^2\right)\left(-2y^2\right)\)
\(=-6x^2y^2+4y^3.\)
Câu 5:
Ta có: \(R=\left(2x-3\right)\left(4+6x\right)-\left(6-3x\right)\left(4x-2\right)\)
\(=\left(8x-12+12x^2-18x\right)-\left(24x-12x^2-12+6x\right)\)
\(=12x^2-10x-12-24x+12x^2+12-6x\)
\(=24x^2-40x.\)
a) \(\left(3x-5\right)\left(3x+5\right)\)
\(=\left(3x\right)^2-5^2\)
\(=9x^2-25\)
b) \(\left(x-2y\right)\left(x+2y\right)\)
\(=x^2-\left(2y\right)^2\)
\(=x^2-4y^2\)
c) \(\left(-x-\dfrac{1}{2}y\right)\left(-x+\dfrac{1}{2}y\right)\)
\(=\left(-x\right)^2-\left(\dfrac{1}{2}y\right)^2\)
\(=x^2-\dfrac{1}{4}y^2\)
`a, (3x-5)(3x+5) = 9x^2 - 25`
`b, (x-2y)(x+2y) = x^2 -4y^2`
`c, (-x-1/2y)(-x+1/2y) = x^2 - 1/4y^2`