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Giải phương trình
e) x4 -4x3-8x2+8x=0
f) 2x2+3xy+y2=0
g) 2x4-x3-9x2+13x-5=0
h) (x+1)(x+3)(x+5)(x+7)+15=0
![](https://rs.olm.vn/images/avt/0.png?1311)
e: =>x(x^3-4x^2-8x+8)=0
=>x[(x^3+8)-4x(x+2)]=0
=>x(x+2)(x^2-2x+4-4x)=0
=>x(x+2)(x^2-6x+4)=0
=>\(x\in\left\{0;-2;3+\sqrt{5};3-\sqrt{5}\right\}\)
g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0
=>(2x+5)(x^3-3x^2+3x-1)=0
=>(2x+5)(x-1)^3=0
=>x=1 hoặc x=-5/2
h: =>(x^2+8x+7)(x^2+8x+15)+15=0
=>(x^2+8x)^2+22(x^2+8x)+120=0
=>(x^2+8x+10)(x^2+8x+12)=0
=>(x^2+8x+10)(x+2)(x+6)=0
=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2x^4-21^3+34x^2+105x+50=0\)
\(\Leftrightarrow2x^4-12x^3-10x^2-9x^3+54x^2+45x-10x^2+60x+50=0\)
\(\Leftrightarrow2x^2\left(x^2-6x-5\right)-9x\left(x^2-6x-5\right)-10\left(x^2-6x-5\right)=0\)
\(\Leftrightarrow\left(x^2-6x-5\right)\left(2x^2-9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-5=0\\2x^2-9x-10=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{14}\\x=3-\sqrt{14}\\x=\dfrac{9+\sqrt{161}}{4}\\x=\dfrac{9-\sqrt{161}}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
9x2 + 4x – 3 – (3x + 2)2 > 0
⇔9x2 + 4x – 3 – (9x2 + 12x + 4) > 0
⇔ 9x2 + 4x – 3 – 9x2 – 12x – 4 > 0
⇔ – 8x > 7 ⇔ x < 7/-8 ⇔ x < -7/8
Tập nghiệm: S = {x|x < -7/8}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
vì x=0 không là nghiệm của pt => chia cả 2 vế cho x2≠0
2x2-7x+9-\(\dfrac{7}{x}\)+\(\dfrac{2}{x^2}\)=0
<=>\(\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+9=0\)
<=>\(2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+9=0\)
đặt \(x+\dfrac{1}{x}\)=y =>\(x^2+\dfrac{1}{x^2}=y^2-2\) ta đc
2(y2-2)-7y+9=0
<=> 2y2-4-7y+9=0
<=>2y2-7y+5=0
<=> 2y2-2y-5y+5=0
<=> (2y2-2y)-(5y-5)=0
<=> 2y(y-1)-5(y-1)=0
<=>(y-1)(2y-5)=0
<=>\(\left\{{}\begin{matrix}y-1=0\\2y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\y=\dfrac{5}{2}\end{matrix}\right.\)
Với y=1 ta có
\(x+\dfrac{1}{x}=1\) =>x2-x+1=0 (vô nghiệm)
Với y=5/2
\(x+\dfrac{1}{x}=\dfrac{5}{2}\) => x=2 và x=\(\dfrac{1}{2}\)
vậy pt có S=\(\left\{2;\dfrac{1}{2}\right\}\)
\(2x^4-7x^3+9x^2-7x+2=0\)
\(\Leftrightarrow2x^4-2x^3-x^3-4x^3+2x^2+x^2+4x^2+2x^2-x-4x-2x+2=0\)
\(\Leftrightarrow\left(2x^4-2x^3+2x^2\right)-\left(x^3-x^2+x\right)-\left(4x^3-4x^2+4x\right)+\left(2x^2-2x+2\right)=0\)
\(\Leftrightarrow2x^2\left(2x^2-2x+2\right)-\dfrac{1}{2}x\left(2x^2-2x+2\right)-2x\left(2x^2-2x+2\right)+\left(2x^2-2x+2\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+2\right)\left(x^2-\dfrac{1}{2}x-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+2\right)\left[x\left(x-\dfrac{1}{2}\right)-2\left(x-\dfrac{1}{2}\right)\right]=0\)
\(\Leftrightarrow\left(2x^2-2x+2\right)\left(x-\dfrac{1}{2}\right)\left(x-2\right)=0\)
Vì: \(2x^2-2x+2=\left(\sqrt{2}x-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}>0\forall x\)
Nên: \(\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy..................
p/s: 1 cách khác :))