(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) x^3 - 9x^2 + 14x = 0`
\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`
\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`
\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`
\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`
\(\Rightarrow\) `x.(x-7)(x-2)=0`
\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)
Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)
`3.x^3 - 5x^2 + 8x - 4 = 0`
\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`
\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`
\(\Rightarrow\) `(x^2 - 4x + 4)(x-1)=0`
\(\Rightarrow\) `(x-2)^2(x-1)=0`
\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)
Vậy \(x\in\left\lbrace2;1\right\rbrace\)
\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)
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
2: \(x^2-2xy+y^2-2x+2y\)
\(=\left(x-y\right)^2-2\left(x-y\right)\)
=(x-y)(x-y-2)
3: \(3x^2-2x-5\)
\(=3x^2-5x+3x-5\)
=x(3x-5)+(3x-5)
=(3x-5)(x+1)
4: \(16-x^2+4xy-4y^2\)
\(=16-\left(x^2-4xy+4y^2\right)\)
\(=4^2-\left(x-2y\right)^2\)
=(4-x+2y)(4+x-2y)
5: \(x^2-2x+1-y^2\)
\(=\left(x-1\right)^2-y^2\)
=(x-1-y)(x-1+y)
6: \(x^2+8x+15\)
\(=x^2+3x+5x+15\)
=x(x+3)+5(x+3)
=(x+3)(x+5)
7: \(\left(x^2+6x+8\right)\left(x^2+14x+48\right)-9\)
=(x+2)(x+4)(x+6)(x+8)-9
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)-9\)
\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384-9\)
\(=\left(x^2+10x\right)^2+15\left(x^2+10x\right)+25\left(x^2+10x\right)+375\)
\(=\left(x^2+10x+25\right)\left(x^2+10x+15\right)=\left(x+5\right)^2\cdot\left(x^2+10x+15\right)\)
8: \(\left(x^2-8x+15\right)\left(x^2-16x+60\right)-24x^2\)
\(=\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+143x^2-24x^2\)
\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+119x^2\)
\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)
\(=\left(x^2-7x+30\right)\left(x-2\right)\left(x-15\right)\)
đầu bài yêu cầu gì thế >?>