(x^2 + 8x+12)(x^2 + 16x+48)-165x^2=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
* \(x^2-8x+12=0\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\) 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 \(x=\dfrac{9}{4};x=\dfrac{-9}{4}\)
+ \(x^2-8x+12=0\)
\(\Rightarrow\left(x^2-2.4x+16\right)-4=0\)
\(\Rightarrow\left(x-4\right)^2-4=0\)
\(\Rightarrow\left(x-4\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
+ \(16x^2-81=0\)
\(\Rightarrow16x^2-9^2=0\)
\(\Rightarrow16x^2=9^2\)
\(\Rightarrow x^2=\dfrac{81}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{81}{16}}\\x=-\sqrt{\dfrac{81}{16}}\end{matrix}\right.\)
\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)
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
\(\Leftrightarrow\Delta\ge0\Leftrightarrow-4A_0+25\ge0\)\(\Leftrightarrow A_0\le\dfrac{25}{4}\)
\(\Rightarrow maxA=\dfrac{25}{4}\)
\(a,9x^2-49=0\)
\(9x^2=49\)
\(x^2=\frac{49}{9}=\frac{7^2}{3^2}=\frac{\left(-7\right)^2}{\left(-3\right)^2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
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 ...
<=> (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= -1
- x+4 = 0 <=> x= -4
đầu bài yêu cầu gì thế >?>