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Câu 4: Không có nghĩa khi x-3=0
=>x=3
Câu 5:
\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)
\(=\left(9-x^2\right)\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)
3, \(x^4-5x^2+4\)
Đặt \(x^2=t\left(t\ge0\right)\)ta có :
\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)
\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
`Answer:`
1. `45+x^3-5x^2-9x`
`=x^3+3x^2-8x^2-24x+15x+45x`
`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`
`=(x+3).(x^2-8x+15)`
`=(x+3).(x^2-5x-3x+15)`
`=(x-3).(x-5).(x-3)`
2. `x^4-2x^3-2x^2-2x-3`
`=x^4+x^3-3x^3+x^2+x-3x-3`
`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`
`=(x+1).(x^3-3x^2+x-3)`
`=(x+1).[x^3 .(x-3).(x-3)]`
`=(x+1).(x-3).(x^2+1)`
3. `x^4-5x^2+4`
`=x^4-x^2-4x^2+4`
`=x^2 .(x^2-1)-4.(x^2-1)`
`=(x^2-1).(x^2-4)`
`=(x-1).(x+1).(x-2).(x+2)`
4. `x^4+64`
`=x^4+16x^2+64-16x^2`
`=(x^2+8)^2-16x^2`
`=(x^2+8-4x).(x^2+8+4x)`
5. `x^5+x^4+1`
`=x^5+x^4+x^3-x^3+1`
`=x^3 .(x^2+x+1)-(x^3-1)`
`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`
`=(x^2+x+1).(x^3-x+1)`
6. `(x^2+2x).(x^2+2x+4)+3`
`=(x^2+2x)^2+4.(x^2+2x)+3`
`=(x^2+2x)^2+x^2+2x+3.(x^2+2x)+3`
`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`
`=(x^2+2x+1).(x^2+2x+3)`
`=(x+1)^2 .(x^2+2x+3)`
7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`
`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`
`=x^6+8x^4+19x^3+30x^2+88x+64`
8. `x^3 .(x^2-7)^2-36x`
`=x[x^2.(x^2-7)^2-36]`
`=x[(x^3-7x)^2-6^2]`
`=x.(x^3-7x-6).(x^3-7x+6)`
`=x.(x^3-6x-x-6).(x^3-x-6x+6)`
`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`
`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`
`=x.(x+1).(x-1).(x^2-3x+2x-6).(x^2+3x-2x-6)`
`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`
`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`
9. `x^5+x+1`
`=x^5-x^2+x^2+x+1`
`=x^2 .(x^3-1)+(x^2+x+1)`
`=x^2 .(x-1).(x^2+x+1)+(x^2+x+1)`
`=(x^2+x+1).(x^3-x^2+1)`
10. `x^8+x^4+1`
`=[(x^4)^2+2x^4+1]-x^4`
`=(x^4+1)^2-(x^2)^2`
`=(x^4-x^2+1).(x^4+x^2+1)`
`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`
`=[(x^2+1)^2-x^2].(x^4-x^2+1)`
`=(x^2-x+1).(x^2+x+1).(x^4-x^2+1)
11. ` x^5-x^4-x^3-x^2-x-2`
`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`
`=x^4 .(x-2)+x^3 ,(x-2)+x^2 .(x-2)+x.(x-2)+(x-2)`
`=(x-2).(x^4+x^3+x^2+x+1)`
12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`
`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`
`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`
`=(x^2-1).(x^7-x^4-x^3+1)`
`=(x-1)(x+1)(x^3-1)(x^4-1)`
`=(x-1)(x+1)(x^2+x+1)(x-1)(x^2-1)(x^2+1)`
`=(x-1)^2 .(x+1)(x^2+x+1)(x-1)(x+1)(x^2+1)`
`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`
13. `(x^2-x)^2-12(x^2-x)+24`
`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`
`=(x^2-x+6)^2-12`
`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề của bạn hơn nhé.
a: \(\dfrac{1}{2a}+\dfrac{2}{3b}\)(ĐKXĐ: a<>0 và b<>0)
\(=\dfrac{1\cdot3b+2\cdot2a}{2a\cdot3b}\)
\(=\dfrac{3b+4a}{6ab}\)
b: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}\)(ĐKXĐ: \(x\notin\left\{1;-1\right\}\))
\(=\dfrac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{-4x}{x^2-1}\)
c: \(\dfrac{x+y}{xy-y}+\dfrac{z}{yz}\)(ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >0\\z< >0\end{matrix}\right.\))
\(=\dfrac{x+y}{y\left(x-1\right)}+\dfrac{1}{y}\)
\(=\dfrac{x+y+x-1}{y\left(x-1\right)}\)
\(=\dfrac{2x+y-1}{y\left(x-1\right)}\)
d: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{2}{x-3}-\dfrac{12}{x^2-9}\)
\(=\dfrac{2}{x-3}-\dfrac{12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: ĐKXĐ: x<>2
\(\dfrac{1}{x-2}+\dfrac{2}{x^2-4x+4}\)
\(=\dfrac{1}{x-2}+\dfrac{2}{\left(x-2\right)^2}\)
\(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2/a/ \(x\left(x+1\right)^2\)
b/ \(\left(y+x\right)\left(y-1\right)\)
c/ \(\left(x+1\right)\left(3x-10\right)\)
d/ \(-3\left(2z+x-y\right)\left(2z+y-x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{-3+6}{9x^2-4}=\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2_{ }-3x+2-3}{\left(3x-2\right)\left(3x+2\right)}=\frac{1}{\left(3x-2\right)\left(3x+2\right)}\)
ý b bạn chỉ cần phân tích mẩu ra nó là 1 hằng đẳng thức rồi lấy mẫu chung là hằng đẩng thưcs đó rồi làm như thường