\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2\\ P=\left(x-y-x-y\right)^2-4x^2\\ P=4y^2-4x^2=4\left(y-x\right)\left(x+y\right)\)

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(\dfrac{3x^2+ax^2+x+a}{x+1}\)

\(=\dfrac{3x^2+3x+ax^2+ax-\left(a+2\right)x-\left(a+2\right)+a+2}{x+1}\)

\(=3x+ax-a-2+\dfrac{a+2}{x+1}\)

Để đây là phép chia hết thì a+2=0

hay a=-2

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

a) x(4x + 2) = 4x² - 14

⇔ 4x² + 2x = 4x² - 14

⇔ 4x² - 4x² + 2x = -14

⇔ 2x = -14

⇔ x = -7

Vậy tập nghiệm S = ......

b) (x² - 9)(2x - 1) = 0

⇔ x² - 9 = 0 hoặc 2x - 1 = 0

⇔ x² = 9 hoặc 2x = 1

⇔ x = 3 hoặc -3 hoặc x = \(\dfrac{1}{2}\)

Vậy .......

c) \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{x^2-4}\) 

⇔ \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{\left(x-2\right)\left(x+2\right)}\)

ĐKXĐ: x - 2 ≠ 0 và x + 2 ≠ 0

Bài 10:

a) (x+2)² -x(x+3) + 5x = -20

=> x² + 4x + 4 - x² - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x³-10x²+5x=0   

=>5x(x²-2x+1)=0

=>5x(x-1)² =0

=> 5x=0 hoặc (x-1)²=0

=>x=0 hoặc x=1

c) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

=> (x² - 1)[(x² - 1)² -  (x⁴ + x² + 1)] = 0

<=> (x² - 1)(x⁴ - 2x² + 1 - x⁴ - x² - 1) = 0

<=>  (x² - 1)(-3x²) = 0

<=> (x² - 1)=0 hoặc (-3x²) =0

<=> x²=1 hoặc x²=0

<=> x=−1;1 hoặc x=0

d)

⇔x=-23/12

Học tốt nhé:333

Bo thi:>

+ đk x > 0 , x khác 1