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\(21x^3-15x^2-6x=0\\ \Leftrightarrow x\left(21x^2-15x-6\right)=0\\ \Leftrightarrow x\left[\left(21x^2-21x\right)+\left(6x-6\right)\right]=0\\ \Leftrightarrow x\left[21x\left(x-1\right)+6\left(x-1\right)\right]=0\\ \Leftrightarrow x\left(x-1\right)\left(21x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-2}{7}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=15x^2\)
\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-5x+6\right)-15x^2=0\) (*)
-Đặt \(t=x^2-5x+6\)
(*) \(\Leftrightarrow t\left(t-2x\right)-15x^2=0\)
\(\Leftrightarrow t^2-2xt-15x^2=0\)
\(\Leftrightarrow t^2-5xt+3xt-15x^2=0\)
\(\Leftrightarrow t\left(t-5x\right)+3x\left(t-5x\right)=0\)
\(\Leftrightarrow\left(t-5x\right)\left(t+3x\right)=0\)
\(\Leftrightarrow t-5x=0\) hay \(t+3x=0\)
\(\Leftrightarrow x^2-5x+6-5x=0\) hay \(x^2-5x+6+3x=0\)
\(\Leftrightarrow x^2-10x+6=0\) hay \(x^2-2x+6=0\)
\(\Leftrightarrow x^2-2.5x+25-19=0\) hay \(\left(x-1\right)^2+5=0\) (pt vô nghiệm)
\(\Leftrightarrow\left(x-5\right)^2-19=0\)
\(\Leftrightarrow\left(x-5-\sqrt{19}\right)\left(x-5+\sqrt{19}\right)=0\)
\(\Leftrightarrow x=5+\sqrt{19}\) hay \(x=5-\sqrt{19}\)
-Vậy \(S=\left\{5+\sqrt{19};5-\sqrt{19}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
g: =>(x-1)(x-2)=0
=>x=1 hoặc x=2
i: \(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
=>x=1 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1. thiếu đề đó bạn ạ
Câu 2:
Ta có: x^3+15x^2+74x+120
=(x^3+4x^2) + (11x^2+44x) + (30x+120)
=(x+4)(x^2+11x+30)
=(x+4)(x+5)(x+6)
Ta có bảng xét dấu
x | -6 | -5 | -4 | ||||
x+4 | - | | | - | | | - | | | + |
x+5 | - | | | - | | | + | | | + |
x+6 | - | | | + | | | + | | | + |
Để (x+4)(x+5)(x+6)<0
Khi có chỉ 1 số âm hoặc cả 3 số âm
<=> x<-6 hoặc -5<x<-4
![](https://rs.olm.vn/images/avt/0.png?1311)
g: \(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>x=1 hoặc x=2
i: \(x^4+x^2+6x-8=0\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x+2\right)\left(x^2-2x+4\right)+x\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
=>x=1 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+7x+12\right)\left(x^2-15x+56\right)=180\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x+4\right)\left(x-7\right)\left(x-8\right)-180=0\)
\(\Leftrightarrow\)\(\left(x^2-4x-21\right)\left(x^2-4x-32\right)-180=0\)
Đặt \(x^2-4x-21=t\) ta có:
\(t\left(t-11\right)-180=0\)
\(\Leftrightarrow\)\(t^2-11t-180=0\)
\(\Leftrightarrow\)\(t^2-20t+9t-180=0\)
\(\Leftrightarrow\)\(\left(t-20\right)\left(t+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}t-20=0\\t+9=0\end{cases}}\)
P/S:đến đây bn thay trở lại rồi tìm x nhé! chúc bn hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)
\(\Leftrightarrow15x-9x+6=45-10x+25\)
\(\Leftrightarrow15x-9x+10x=45+25-6\)
\(\Leftrightarrow16x=64\)
\(\Leftrightarrow x=4\)
b) \(x^2-9+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)
\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)
a) 15x - 3(3x - 2) = 45 - 5(2x - 5)
\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25
\(\Leftrightarrow\) 6x + 10x = 70 - 6
\(\Leftrightarrow\) 16x = 64
\(\Leftrightarrow\) x = 4
Vậy.......................
b) x2 - 9 + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)
Vậy........................
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4
\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0
\(\Leftrightarrow\) x2 - 6x = 0
\(\Leftrightarrow\) x(x - 6) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)
Vậy...............
![](https://rs.olm.vn/images/avt/0.png?1311)
+ Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow\left(x^3+5x^2\right)-\left(6x^2+30x\right)+\left(9x+45\right)=0\)
\(\Leftrightarrow x^2.\left(x+5\right)-6x.\left(x+6\right)+9.\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right).\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right).\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{-5,3\right\}\)
+ Ta có: \(\left(x^2-2x+1\right)-9=0\)
\(\Leftrightarrow x^2-2x+1-9=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(2x-8\right)=0\)
\(\Leftrightarrow x.\left(x-4\right)+2.\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2,4\right\}\)
+ Ta có: \(x.\left(x-2\right)=-x+12\)
\(\Leftrightarrow x^2-2x+x-12=0\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(3x-12\right)=0\)
\(\Leftrightarrow x.\left(x-4\right)+3.\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right).\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(TM\right)\\x=-3\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{-3,4\right\}\)
a) X^3-x^2-21x+45=0
x^3-3x^2+2x^2-6x-15x+45=0
x^2(x-3)+2x(x-3)-15(x-3)=0
(x-3)(x^2+2x-15)=0
(x-3)(x^2-3x+5x-15)=0
(x-3)[x(x-3)+5(x-3)]=0
(x-3)^2(x+5)=0
<=> x=3 hoặc x=-5
Câu 2 đề ko rõ lắm bn sửa lại đề để mk giải hộ nha
Bích Ngọc bạn xem lời giải dưới đây nhé :
X^3-x^2-21x+45=0\(\Leftrightarrow\)(x+5)(x^2-6x+9)=0
\(\Leftrightarrow\)(x+5)(x-3)^2=0
Rồi đó tới đây bạn tự tìm x nhé!