(x^2 + 8x+12)(x^2 + 16x+48)-165x^2=0
giải phương trình: (x^2+8x+12)(x^2+16x+48)-165x^2=0
(x^2+8x+16)(x^2+16x+48) -165x^2=0
Tìm X
x2-8x+12=0
x2+5x-14=0
16x2-81=0
* \(x^2-8x+12=0\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\) vậy \(x=2;x=6\)
* \(x^2+5x-14=0\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\) vậy \(x=-7;x=2\)
* \(16x^2-81=0\Leftrightarrow16\left(x^2-\dfrac{81}{16}\right)=0\Leftrightarrow x^2-\dfrac{81}{16}=0\)
\(\Leftrightarrow x^2=\dfrac{81}{16}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{81}{16}}\\x=-\sqrt{\dfrac{81}{16}}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\) vậy \(x=\dfrac{9}{4};x=\dfrac{-9}{4}\)
+ \(x^2-8x+12=0\)
\(\Rightarrow\left(x^2-2.4x+16\right)-4=0\)
\(\Rightarrow\left(x-4\right)^2-4=0\)
\(\Rightarrow\left(x-4\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
+ \(16x^2-81=0\)
\(\Rightarrow16x^2-9^2=0\)
\(\Rightarrow16x^2=9^2\)
\(\Rightarrow x^2=\dfrac{81}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{81}{16}}\\x=-\sqrt{\dfrac{81}{16}}\end{matrix}\right.\)
3. x^3-5x^2+8x-4=0
4. c) x^3 - 3x^2 -16x+48=0
cho x>0 tim Max \(A=\dfrac{1200x\left(12+x\right)}{\left(48+16x\right)^2}\)
\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)
Gọi \(A_0\in MGT\) của A
\(\Rightarrow A_0=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\) có nghiệm
\(\Rightarrow A_0\left(12+4x\right)^2=75x\left(12+x\right)\)
\(\Leftrightarrow x^2\left(16A_0-75\right)+x\left(96A_0-900\right)+144A_0=0\) có nghiệm
\(\Leftrightarrow\Delta\ge0\Leftrightarrow-4A_0+25\ge0\)\(\Leftrightarrow A_0\le\dfrac{25}{4}\)
\(\Rightarrow maxA=\dfrac{25}{4}\)
a) 9x2 -49 = 0
b) 5x2 -4(x2-2x+1)-5=0
c) x3 -16x=0
d) (x2-9)2-(x-3)2=0
f)x(8x-2)-8x2=12=0
\(a,9x^2-49=0\)
\(9x^2=49\)
\(x^2=\frac{49}{9}=\frac{7^2}{3^2}=\frac{\left(-7\right)^2}{\left(-3\right)^2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
vậy ...
\(c,x^3-16x=0\)
\(x.\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=4,x=-4\end{cases}}\)
vậy ...
Phân tích đa thức thành nhân tử
a, x2 - 2xy + 4y2 -2x + 4y - 35
b, (x2 + 2)2 + 9x2 + 16x + 20
c, x(x+4)(x+6)(x+10) +128
d, (x+a)(x+2a)(x+3a)(x+4a) +a4
f, (x+12)(x+6)(x+4)(x+2) - 165x2
giải phương trình : (x^2+x+4)^2+8x(x^2+x+4)+16x^2=0
<=> (x2 +x +4)2 + 2 . 4x(x2+ x + 4) + (4x)2 = 0
<=> ( x2 + x+ 4 +4x )2 = 0
<=> [(x2 + x) + (4 +4x)] =0
<=> [x(x+1) + 4(1+x)] =0
<=> (x+1) + (x+4) =0
x+1 = 0 <=> x= -1x+4 = 0 <=> x= -42)x^2-2xy+y^2-2x+2y
3)3x^2-2x-5
4)16-x^2+4xy-4x^2
5)x^2-2x+1-y^2
6)x^2+8x+15
7)(x^2+6x+8)(x^2+14x+48)-9
8)(x^2-8x+15)(x^2-16x+60)-24x^2
9)x^5+x^4+1
10)x^4-x^3-10x^2+2x+4