D.\(x^2+5x+9< 0\)

\(x^2+5x+9=\left(x^2+2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right)-\left(\dfrac{5}{2}\right)^2+9=\left(x+\dfrac{5}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

Mà \(x^2+5x+9< 0\)

--> pt vô nghiệm

Ta có: \(x^3-5x^2+6x=0\)

\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={0;2;3}

\(x^3-5x^2+6x=0\)

\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)

\(S=\left\{0,2,3\right\}\)

x^3 -5x^2 + 6x=0

\(< =>x\left(x^2-5x+6\right)=0\)

\(< =>x=0;x^2-5x+6=0\)

+/ \(x^2-5x+6=0\)

\(< =>x^2-2x-3x+6=0\)

\(< =>\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(< =>x\left(x-2\right)-3\left(x-2\right)=0\)

\(< =>\left(x-2\right)\left(x-3\right)=0\)

=> \(x-2=0;x-3=0\)

=> \(x=2;3\)

Vayayj tập nghiệm của......

D

D

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

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

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

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

\(\left(\frac{x-1}{x+2}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)^2+3\left(\frac{x+1}{x-2}\right)^2=0\left(1\right)\)

\(ĐKXĐ:x\ne\pm2\)

Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)

=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)

\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)

=>  \(b=0;a=0\)

Bạn cùng trường :">

a) pt<=> \(\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)

     <=>\(\left|x-2\right|+\left|x-3\right|=1\)

đến đây chia 3 trường hợp để phá trị tuyệt đối là ra 

b) \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)

<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\)

câu này cũng tương tự câu a nha

\(\frac{5}{\sqrt{x^2}+1}\)hay\(\frac{5}{\sqrt{x^2+1}}\)v
b)
Đặt \(\sqrt{x-2}=a\); \(\sqrt{4-x}=b\)
Ta có hpt:
\(\hept{\begin{cases}a+b=-a^2b^2+3\\a^2+b^2=2\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=-a^2b^2+3\\\left(a+b\right)^2-2ab-2=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2\\\left(-a^2b^2+3\right)^2-2ab-2=0\end{cases}}\)
Đặt ab=t rồi giải hệ nhé bạn

Phần b cách ngắn hơn nè:
\(\sqrt{x-2}-1+\sqrt{4-x}-1=x^2-6x+9\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2}\right)^2-1}{\sqrt{x-2}+1}+\frac{\left(\sqrt{4-x}\right)^2-1}{\sqrt{4-x}+1}=\left(x-3\right)^2\)
\(\Leftrightarrow\frac{x-3}{\sqrt{x-2}+1}+\frac{3-x}{\sqrt{4-x}+1}=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}-x+3\right)=0\)
\(\Rightarrow x=3\)
 

\(\sqrt{x^2-1}+\sqrt{x^2-9}=8\)

\(\Leftrightarrow\sqrt{\sqrt{x^2-8}+\sqrt{x^2-9}}+\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{7x+2\sqrt{x}}+9\sqrt{x^2-9}=0\)

\(\Leftrightarrow7x\sqrt{x^2+8}+2\sqrt{x^2-6}=6\)

\(\Leftrightarrow x^2-5=0\)

\(\Leftrightarrow x=\pm\sqrt{5}\)