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Ta có : a3 + b3 = (a + b)(a - ab + b)
Thay ab = 4 và a + b = 5
=> a3 + b3 = 5(5 - 4)
=> a3 + b3 = 5
Vậy a3 + b3 = 5
Bài 1:
a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)
b) \(\left(2x-3\right).\left(x+2\right)=2x^2+4x-3x-6=2x^2-x-6\)
Bài 2:
a) \(x^3-2x^2+x=x.\left(x^2-2x+1\right)=x.\left(x-1\right)^2\)
b) \(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x.\left(x-y\right)+2.\left(x-y\right)=\left(x-y\right).\left(x+2\right)\)
c) Đề sai.
phân tích n^3 + 3n^2 + 2n thảnh n.(n+1).(n+2) chia hết cho 6 vì chia hết cho 2 và 3 chia hết cho 15 là chia hết cho 3 với 5 nha
a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)
\(=x\left(x+y-2z\right)\left(x+y+2z\right)\)
b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)
c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)
d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
e)\(x\left(2-x\right)-x+2=x\left(2-x\right)+\left(2-x\right)=\left(2-x\right)\left(x+1\right)\)
f)\(2x^2+3x-2=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g)\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)\)\(=\left(x-3\right)^2\left(x+3\right)\)
B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)
1.x2-9
= (x-3)(x+3)
2. -2x2+2x+12
= -2x2+6x-4x+12
= -2x(x+2)+6(x+2)
= (x+2)(-2x+6)
4. -2x2+2x+24
= -2x2+8x-6x+24
= -2x(x+3)+8(x+3)
= (x+3)(-2x+8)
6. x2-5x+4
= x2-4x-x+4
= x(x-1) -4(x-1)
= (x-1)(x-4)
8. x2-7x+6
= x2-6x-x+6
= x(x-1)-6(x-1)
= (x-1)(x-6)
9. x2+5x+4
= x2+4x+x+4
= x(x+1)+4(x+1)
=(x+1)(x+4)
10. x2+7x+6
= x2 +x+6x+6
= x(x+1)+6(x+1)
= (x+6)(x+1)
K nhé
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
2) \(1-9x^2=\left(1-3x\right)\left(1+3x\right)\)
3) \(\frac{x^2}{9}-\frac{y^2}{16}=\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)\)
4) \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)
5) \(\left(a-b\right)^2-1=\left(a-b+1\right)\left(a-b-1\right)\)
6) \(4-\left(a-b\right)^2=\left(2-a+b\right)\left(2+a-b\right)\)
7) \(\left(x-y\right)^2-\left(m+n\right)^2=\left(x-y-m-n\right)\left(x-y+m+n\right)\)
8) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\)
\(=\left[3\left(x+y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x-y\right)\right]=5\left(x+y\right)\left(x-y\right)\)
9) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
10) \(\left(x^4+2x^2+1\right)=\left(x^2+1\right)^2\)
11) \(\left(a^4+4-4x^2\right)=\left(a^2-2\right)^2\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`3x^2 + 6xy + 3y^2 - 3z`
`= 3*x^2 + 3*2xy + 3y^2 - 3z`
`= 3(x^2 + 2xy + y^2 - z)`
`b,`
`x^3 + x^2y - x^2z - xyz`
`= x(x + y)(x-z)`