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*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
1/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
2/ \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))
\(\Leftrightarrow x=-1\).
3/ \(x^3-6x^2+8x=0\)
\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)
4/ \(x^4-8x^3-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)
Ta có : \(x^2-2x-1=0
\)
\(\Leftrightarrow \)\((x-1)^2=2\)
\(\Leftrightarrow
\)\(\left[\begin{array}{}
x-1=\sqrt{2}\\
x-1=-\sqrt{2}
\end{array} \right.\)
Đặt P = \(\dfrac{x^6-6x^5+12x^4-8x^3+2015}{x^6-8x^3-12x^2+6x+2015}\)
=\(\dfrac{(x^6-2x^5-x^4)-(4x^5-8x^4-4x^3)+(5x^4-10x^3-5x^2)-(2x^3-4x^2-2x)+(x^2-2x-1)+2016}
{(x^6-2x^5-x^4)+(2x^5-4x^4-2x^3)+(5x^4-10x^3-5x^2)+(4x^3-8x^2-4x)+(x^2-2x-1)+12x+2016}\)
=\(\dfrac{x^4(x^2-2x-1)-4x^3(x^2-2x-1)+5x^2(x^2-2x-1)-2x(x^2-2x-1)+(x^2-2x-1)+2016}
{x^4(x^2-2x-1)+2x^3(x^2-2x-1)+5x^2(x^2-2x-1)+4x(x^2-2x-1)+(x^2-2x-1)+12x+2016}\)
=\(\dfrac{2016}{12x + 2016}\)
=\(\dfrac{2016}{12(x+1)+2004}\)
=\(\dfrac{168}{x+1+167}\)
=\(\left[\begin{array}{}
\dfrac{168}{\sqrt{2}+167}\\
\dfrac{168}{-\sqrt{2}+167}
\end{array} \right.\)
Chú thích: Hình như mẫu là \(-6x\) chứ không phải \(6x
\) bạn ạ. Hay là mình phân tích sai thì cho mình xin lỗi nhé.
a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)
a) ĐKXĐ: \(x\notin\left\{\frac{1}{2};\frac{-1}{2}\right\}\)
Ta có: \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)
\(\Leftrightarrow\frac{8x+1}{4\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}+\frac{8x^2}{3\left(4x^2-1\right)}\)
\(\Leftrightarrow\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}=\frac{2x\cdot4\cdot\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}+\frac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(3\left(8x+1\right)\left(2x-1\right)=8x\left(2x+1\right)+32x^2\)
\(\Leftrightarrow3\left(16x^2-8x+2x-1\right)=16x^2+8x+32x^2\)
\(\Leftrightarrow3\left(16x^2-6x-1\right)=48x^2+8x\)
\(\Leftrightarrow48x^2-18x-3-48x^2-8x=0\)
\(\Leftrightarrow-26x-3=0\)
\(\Leftrightarrow-26x=3\)
hay \(x=-\frac{3}{26}\)
Vậy: \(S=\left\{-\frac{3}{26}\right\}\)
b) Ta có: \(\left(x-2\right)\left(x-3\right)< \left(x-4\right)^2-2\left(x+3\right)\)
\(\Leftrightarrow x^2-5x+6< x^2-8x+16-2x-6\)
\(\Leftrightarrow x^2-5x+6< x^2-10x+10\)
\(\Leftrightarrow x^2-5x+6-x^2+10x-10< 0\)
\(\Leftrightarrow5x-4< 0\)
\(\Leftrightarrow5x< 4\)
hay \(x< \frac{4}{5}\)
Vậy: S={x|\(x< \frac{4}{5}\)}
a: \(=x\left(x^3+4x^2-8x+8\right)\)
b: \(=6x^2+8x-3x-4=\left(3x+4\right)\left(2x-1\right)\)
a) x3 - 4x2 - 8x + 8
= x3 + 2x2 - 6x2 - 12x + 4x + 8
= x2(x + 2) - 6x(x + 2) + 4(x + 2)
= (x + 2)(x2 - 6x + 4)
b) 1 + 6x - 6x2 - x3
= -x3 + x2 - 7x2 + 7x - x + 1
= -x2(x - 1) - 7x(x - 1) - (x - 1)
= -(x - 1)(x2 + 7x + 1)
c) 6x3 - x2 - 486x + 81
= 6x2(x - 1/6) - 486(x - 1/6)
= (x - 1/6)(6x2 - 486)
= 6(x - 1/6)(x2 - 81)
= 6(x - 1/6)(x - 9)(x + 9)
câu a:
\(8x^2-6x+3-2x=\left(2x-1\right)\sqrt{8x^2-6x+3}\)
đặt \(t=\sqrt{8x^2-6x+3}\Leftrightarrow t^2=8x^2-6x+3\)phương trình trở thành
\(t^2-2x=\left(2x-1\right)t\Leftrightarrow t^2-\left(2x-1\right)t-2x=0\)
có \(\Delta=\left(2x-1\right)^2+8x=\left(2x+1\right)^2\Rightarrow\orbr{\begin{cases}t=-1\\t=2x\end{cases}}\)
- \(t=-1\Rightarrow8x^2-6x+3=1\Leftrightarrow8x^2-6x+2=0VN\)
- \(t=2x\Rightarrow8x^2-6x+3=4x^2\Leftrightarrow4x^2-6x+3=0VN\)
Câu b:
Đặt \(t=\sqrt{x^2+1}\Leftrightarrow t^2=x^2+1\left(t>0\right)\)
PT\(\Leftrightarrow t^2-\left(x+3\right)t+3x=0\)
có :\(\Delta=\left(x+3\right)^2-4.3x=\left(x-3\right)^2\Rightarrow\orbr{\begin{cases}t=3\\t=x\end{cases}}\)
- \(t=3\Rightarrow9=x^2+1\Leftrightarrow x^2=8\Leftrightarrow\orbr{\begin{cases}x=2\sqrt{2}\\x=-2\sqrt{2}\end{cases}}\)
- \(t=x\Leftrightarrow x^2=x^2+1VN\)
\(\frac{x-2}{8x^2-8x}-\frac{x+1}{6x^3-6x^2}=\frac{x-1}{8x\left(x-1\right)}-\frac{x+1}{6x^2\left(x-1\right)}\)
\(=\frac{3x\left(x-1\right)}{24x^2\left(x-1\right)}-\frac{\left(x+1\right)4}{24x^2\left(x-1\right)}=\frac{3x^2-3x-4x-4}{24x^2\left(x-1\right)}\)
\(=\frac{3x^2-7x-4}{24x^2\left(x-1\right)}\)