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THay x =2 vào phương trình:
8m - 8 +12 +m^2+4 = 0
m^2+ 8m + 8 =0
(m+4)^2 =8
m+4 = căn 8 hay m+4 = âm căn 8
m = căn 8 -4 hay m = âm căn 8 -4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(x^3-x=0\Rightarrow x\left(x^2-1\right)=0\Rightarrow x\left(x+1\right)\left(x-1\right)=0\)
b,\(x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+1\right)=0\)
c,\(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(x^3-x=0\)
\(\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
x=0 hoặc x-1=0=> x=1 hoặc x+1=0 => x=-1
\(x^2-2x+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(x^2-6x+8=0\)
\(\Leftrightarrow x^2-2x-4x+8=0\)
\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+y^2+13z^2-4yz-6x+9=0\)
\(\Leftrightarrow\left(2x^2-6x+\dfrac{9}{2}\right)+\left(y^2-4yz+4z^2\right)+9z^2+\dfrac{9}{2}=0\)
\(\Leftrightarrow2\left(x^2-3x-\dfrac{9}{4}\right)+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}=0\)
\(\Leftrightarrow2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}=0\)
Dễ thấy: \(2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2\ge0\forall x,y,z\)
\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x,y,z\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}2\left(x-\dfrac{3}{2}\right)^2=0\\\left(y-2z\right)^2=0\\9z^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{2}=0\\y=2z\\z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=0\\z=0\end{matrix}\right.\)
Khi đó \(P=\dfrac{2\cdot\dfrac{3}{2}\cdot0+\dfrac{3}{2}\cdot0-\left(\dfrac{3}{2}\right)^2-2\cdot0^2-0\cdot0}{\left(\dfrac{3}{2}\right)^2-0^2}=-1\)
Đệch, theo đề bài của bn thì Thắng làm đúng òi
Hình như đề thiếu -6xz mới ra -4/5
![](https://rs.olm.vn/images/avt/0.png?1311)
B1.a/ (x-2)(x^2+2x+2)
b/ (x+1)(x+5)(x+2)
c/ (x+1)(x^2+2x+4)
B2.
1a) x3 - 2x - 4 = 0
<=> (x3 - 4x) + (2x - 4) = 0
<=> x(x2 - 4) + 2(x - 2) = 0
<=> x(x - 2)(x + 2) + 2(x - 2) = 0
<=> (x - 2)(x2 + 2x + 2) = 0
<=> x - 2 = 0 (vì x2 + 2x + 2 \(\ne\)0)
<=> x = 2
Vậy S = {2}
b) x3 + 8x2 + 17x + 10 = 0
<=> (x3 + 5x2) + (3x2 + 15x) + (2x + 10) = 0
<=> x2(x + 5) + 3x(x + 5) + 2(x + 5) = 0
<=> (x2 + 3x + 2)(x + 5) = 0
<=> (x2 + x + 2x + 2)(x + 5) = 0
<=> (x + 1)(x + 2)(x + 5) = 0
<=> x + 1 = 0 hoặc x + 2 = 0 hoặc x + 5 = 0
<=> x = -1 hoặc x = -2 hoặc x = -5
Vậy S = {-1; -2; -5}
c) x3 + 3x2 + 6x + 4 = 0
<=> (x3 + x2) + (2x2 + 2x) + (4x + 4) = 0
<=> x2(x + 1) + 2x(x + 1) + 4(x + 2) = 0
<=> (x2 + 2x + 4)(x + 2) = 0
<=> x + 2 = 0
<=> x = -2
Vậy S = {-2}
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c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-6x^3+7x^2+6x-8=0\)
\(\Leftrightarrow x^4-4x^3-2x^3+8x^2-x^2+4x+2x-8=0\)
\(\Leftrightarrow x^3\left(x-4\right)-2x^2\left(x-4\right)-x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3-2x^2-x+2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[x^2\left(x-2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{-1;1;2;4\right\}\)
Vậy S={-1;1;2;4}
(x2 - 6x + 9) - 4 = 0
<=> (x - 3)2 - 22 = 0
<=> (x - 5).(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy...
(x2 - 6x + 9) - 4 = 0
x2 - 6x + 5 = 0
=> x = 5 hoặc 1
K chắc nhé :>