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a) Ta có: \(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
b) Ta có: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
c) Ta có: \(6x^2-11x-16\)
\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)
\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)
\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)
\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)
d) Ta có: \(4x^2-8x-5\)
\(=4x^2-10x+2x-5\)
\(=2x\left(2x-5\right)+\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+1\right)\)
e) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
g) Ta có: \(x^3+9x^2+23x+15\)
\(=x^3+x^2+8x^2+8x+15x+15\)
\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+8x+15\right)\)
\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
h) Ta có: \(2x^4-x^3-9x^2+13x\)
\(=x\left(2x^3-x^2-9x+13\right)\)
i) Ta có: \(x^4+2x^3-16x^2-2x+15\)
\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)
\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)
\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)
\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)
\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)
a)
\(x^3+6x^2+11x+6=(x^3-x)+(6x^2+12x+6)\)
\(=x(x^2-1)+5(x^2+2x+1)\)
\(=x(x-1)(x+1)+6(x+1)^2\)
\(=(x+1)[x(x-1)+6(x+1)]=(x+1)(x^2+5x+6)\)
\(=(x+1)(x^2+2x+3x+6)\)
\(=(x+1)[x(x+2)+3(x+2)]=(x+1)(x+2)(x+3)\)
b) \(x^3+6x^2-13x-42\)
\(=x^3+2x^2+4x^2+8x-21x-42\)
\(=x^2(x+2)+4x(x+2)-21(x+2)\)
\(=(x+2)(x^2+4x-21)\)
\(=(x+2)[x^2-3x+7x-21)\)
\(=(x+2)(x+7)(x-3)\)
c)
\(x^3-5x^2+8x-4=(x^3-x^2)-4x^2+8x-4\)
\(=x^2(x-1)-4(x^2-2x+1)\)
\(=x^2(x-1)-4(x-1)^2\)
\(=(x-1)[x^2-4(x-1)]=(x-1)(x^2-4x+4)\)
\(=(x-1)(x-2)^2\)
d) \(2x^3-x^2+3x+6\)
\(=2x^3+2x^2-3x^2+3x+6\)
\(=2x^2(x+1)-3(x^2-x-2)\)
\(=2x^2(x+1)-3[x^2+x-2x-2]\)
\(=2x^2(x+1)-3[x(x+1)-2(x+1)]\)
\(=2x^2(x+1)-3(x+1)(x-2)\)
\(=(x+1)(2x^2-3x+6)\)
a) \(\dfrac{2x+3}{x-5}=\dfrac{2\left(x-5\right)+13}{x-5}=2+\dfrac{13}{x-5}\)
Để \(2+\dfrac{13}{x-5}\in Z\)
thì \(\dfrac{13}{x-5}\in Z\Rightarrow13⋮x-5\)
\(\Rightarrow x-5\inƯ\left(13\right)\)
\(\Rightarrow x-5\in\left\{\pm1;\pm13\right\}\)
Xét các trường hợp...
b) \(\dfrac{x^3-x^2+2}{x-1}=\dfrac{x^2\left(x-1\right)+2}{x-1}=x^2+\dfrac{2}{x-1}\)
Tương tự câu a)
c) \(\dfrac{x^3-2x^2+4}{x-2}=\dfrac{x^2\left(x-2\right)+4}{x-2}=x^2+\dfrac{4}{x-2}\)
...
d) \(\dfrac{2x^3+x^2+2x+2}{2x+1}=\dfrac{x^2\left(2x+1\right)+2x+2}{2x+1}=x^2+\dfrac{2x+2}{2x+1}\)
Khi đó lí luận cho \(2x+2⋮2x+1\)
\(\Rightarrow\left(2x+1\right)+1⋮2x+1\)
\(\Rightarrow1⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(1\right)\)
...
e) \(\dfrac{3x^3-7x^2+11x-1}{3x-1}=\dfrac{x^2\left(3x-1\right)-2x\left(3x-1\right)+3\left(3x-1\right)+2}{3x-1}\)
\(=\dfrac{\left(x^2-2x+3\right)\left(3x-1\right)+2}{3x-1}=\left(x^2-2x+3\right)+\dfrac{2}{3x-1}\)
...
f) \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}=\dfrac{\left(x^2\right)^2-4^2}{\left(x-2\right)^2\left(x^2+4\right)}\)
\(=\dfrac{\left(x^2-4\right)\left(x^2+4\right)}{\left(x-2\right)^2\left(x^2+4\right)}=\dfrac{x^2-4}{\left(x-2\right)^2}=\dfrac{x+2}{x-2}=\dfrac{\left(x-2\right)+4}{x-2}=1+\dfrac{4}{x-2}\)
....
a) \(x^3-5x^2+8x-4\)
= \(x^3-x^2-4x^2+4x+4x-4\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
b) \(x^3-9x^2+6x+16\)
= \(\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
c) \(x^3+2x-3\)
= \(x^3-x^2+x^2-x+3x-3\)
= \(x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+x+3\right)\)
d) \(2x^3-12x^2+17x-2\)
= \(2x^3-4x^2-8x^2+16x+x-2\)
= \(2x^2\left(x-2\right)-8x\left(x-2\right)+\left(x-2\right)\)
= \(\left(x-2\right)\left(2x^2-8x+1\right)\)
e) \(x^3-5x^2+3x+9\)
= \(x^3+x^2-6x^2-6x+9x+9\)
= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)
f) \(x^3-8x^2+17x+10\)
Câu này có vẻ sai đề, nghiệm cực kì khủng bố @@
g) \(x^3-2x-4\)
= \(x^3-2x^2+2x^2-4x+2x-4\)
= \(x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
= \(\left(x-2\right)\left(x^2+2x+2\right)\)
h) \(x^3+x^2+4\)
= \(x^3+2x^2-x^2+4\)
= \(x^2\left(x+2\right)-\left(x-2\right)\left(x+2\right)\)
= \(\left(x+2\right)\left(x^2-x+2\right)\)
i) \(x^3-7x+6\)
= \(\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
a)\(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b)\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c)\(x^2-5x-24=x^2+3x-8x-24=x\left(x+3\right)-8\left(x+3\right)=\left(x-8\right)\left(x+3\right)\)
d)\(3x^2-16x+5=3x^2-x-15x+5=x\left(3x-1\right)-5\left(3x-1\right)=\left(x-5\right)\left(3x-1\right)\)
e)\(8x^2+30x+7=8x^2+28x+2x+7=4x\left(2x+7\right)+\left(2x+7\right)=\left(4x+1\right)\left(2x+7\right)\)
g)\(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h)\(6x^2-7x+3=6x^2-9x-2x+3=3x\left(2x-3\right)-\left(2x-3\right)=\left(3x-1\right)\left(2x-3\right)\)
i)\(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k)\(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
a ) \(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b ) \(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c ) \(x^2-5x-24=x^2-3x+8x-24=x\left(x-3\right)+8\left(x-3\right)=\left(x+8\right)\left(x-3\right)\)
d ) \(3x^2-16x+5=3x^2-15x-x+5=3x\left(x-5\right)-\left(x-5\right)=\left(3x-1\right)\left(x-5\right)\)
e ) \(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)
g ) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h ) \(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(3x+4\right)\left(2x-5\right)\)
i ) \(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k ) \(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
a)\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b)\(\left(x-3\right)\left(x-7\right)\left(x+2\right)\)
c)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)\left(x+1\right)\)
d)\(\left(x+5\right)\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a)
\(x^3-7x-6=x^3-x-6x-6\)
\(=x(x^2-1)-6(x+1)\)
\(=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]\)
\(=(x+1)(x^2-x-6)=(x+1)[x^2-3x+2x-6]\)
\(=(x+1)[x(x-3)+2(x-3)]=(x+1)(x+2)(x-3)\)
b) \(x^3-6x^2+8x\)
\(=x(x^2-6x+8)\)
\(=x(x^2-4x-2x+8)\)
\(=x[x(x-4)-2(x-4)]=x(x-2)(x-4)\)
c) \(x^4+2x^3-16x^2-2x+15\)
\(=(x^4+2x^3-x^2-2x)-15x^2+15\)
\(=[(x^4-x^2)+(2x^3-2x)]-15(x^2-1)\)
\(=[x^2(x^2-1)+2x(x^2-1)]-15(x^2-1)\)
\(=(x^2-1)(x^2+2x)-15(x^2-1)=(x^2-1)(x^2+2x-15)\)
\(=(x^2-1)(x^2-3x+5x-15)=(x^2-1)[x(x-3)+5(x-3)]\)
\(=(x^2-1)(x+5)(x-3)=(x-1)(x+1)(x+5)(x-3)\)
d)
\(x^3-11x^2+30x=x(x^2-11x+30)\)
\(=x(x^2-5x-6x+30)\)
\(=x[x(x-5)-6(x-5)]=x(x-6)(x-5)\)