/X/x8/3=5 1/3
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b) \(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
\(\dfrac{2\cdot3\cdot8}{4\cdot5\cdot6\cdot7}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{8}{35}=\dfrac{2}{35}\)
`x^8+36x^4=0`
`<=>x^4(x^4+36)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^4=0\\x^4+36=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^4=-36\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)
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`(x-5)^3 -x+5=0`
`<=> (x-5)^3 -(x-5)=0`
`<=> (x-5) [(x-5)^2 -1]=0`
`<=> (x-5)(x-5-1)(x-5+1)=0`
`<=>(x-5)(x-6)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
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`5(x-2)-x^2+4=0`
`<=>5(x-2)-(x^2-4)=0`
`<=>5(x-2)-(x-2)(x+2)=0`
`<=>(x-2)(5-x-2)=0`
`<=>(x-2)(-x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
|2x+3|=5
=> 2x+3 thuộc -5 và 5
Th1:2x+3=5
=>x=1
Th2: 2x+3=-5
x=-4
Vậy x thuộc 1 và -4
dấu " \(.\) " là dấu nhân
\(\left\{\left[27.11-\left(x+12\right).8\right].3-79\right\}:5=100\)
=>\(\left[297-8.x+12.8\right].3-79=100.5\)
=> \(\left[297-8.x+96\right].3-79=500\)
=> \(\left[297-8.x+96\right].3=500+79\)
=> \(\left[\left(297+96\right)-8.x\right].3=579\)
=> \(393-8.x=579:3\)
=> \(393-8.x=193\)
=> \(8.x=393-193\)
=> \(8.x=200=>x=200:8=25\)
Vậy \(x=25\)
Chúc bạn Hk tốt!!!!
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!