\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)

\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)

\(x^4-8x^3+21x^2-24x+9=0\)

\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))

\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

1)x^4+x^2-6x+1=0>>>x^4+4x^2+4-3x^2-6x-3=0>>>(x^2+2)^2=3(x-1)^2.

>>Sau đó giải bt.

2)Đặt x^2-x+1=a;x+1=b thì:x^3+1=ab.

Pt:2a+5b^2+14ab=0(tự giải nha)

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

            \(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)

           \(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)

Bài 1 : 

\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)

Vì bạn không viết công thức toán nên không rõ tử là x-3 hay 3, x-2 hay 2. Tương tự mẫu cũng vậy. 

Bạn nên viết đề bằng công thức toán để được hỗ trợ tốt hơn.

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

1, \(x^4-19x^2-10x+8=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3-4x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)\left(x^2-5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\\x^2-5x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x_1=-4\\x_2=-1\end{matrix}\right.\)

hoặc \(x^2-5x+2=0\)

\(\Rightarrow\Delta=17\left(CT:b^2-4ac\right)\)

\(\Rightarrow\left[{}\begin{matrix}x_3=\dfrac{5+\sqrt{17}}{2}\\x_4=\dfrac{5-\sqrt{17}}{2}\end{matrix}\right.\)

Vậy pt có 4 n^o là...........

(1)-a)Với mọi x, ta luôn có: \(\left(x+1\right)^2+3>0\Leftrightarrow x^2+1+2x+3>0\Leftrightarrow x^2+2x+4>0\)

            \(\sqrt{x^2+2x+4}=2\Leftrightarrow x^2+2x+4=2^2=4\)

                                           \(\Leftrightarrow x^2+2x=0\\\Leftrightarrow\left(x+2\right)x=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\Leftrightarrow x=-2\\x=0\end{matrix}\right. \)

        ➤\(x\in\left\{-2;0\right\}\)

b) \(\left\{{}\begin{matrix}x+2y-1=0\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=1\\4x+2y=10\end{matrix}\right.\)

                                  \(\Leftrightarrow\left\{{}\begin{matrix}2y=1-x\\3x=9\Leftrightarrow x=\dfrac{9}{3}=3\end{matrix}\right.\)

Do \(x=3\Leftrightarrow1-x=1-3=-2\) nên ta có: \(2y=1-x=-2\Leftrightarrow y=\dfrac{-2}{2}=-1\)

➤\(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

(2): +)ĐK để 2 hàm số cắt nhau là: \(2a\ne1\Leftrightarrow a\ne\dfrac{1}{2}\Leftrightarrow a\ne0,5\) 

Ta có hệ phương trình sau: \(\left\{{}\begin{matrix}y=2ax+a+1\\y=x+2\end{matrix}\right.\)

➢Do đó, ta có: \(2ax+a+1=x+2\Leftrightarrow2ax+a-x=2-1=1\)

\(=x^2-6x+8-x^2+2x-1=-4x+7\)

chia 3 khoảng để xét dấu đi bạn