a: 11x+4=-3/2

=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)

=>\(x=-\dfrac{1}{2}\)

b: \(x^2-9+2\left(x-3\right)=0\)

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

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

=>(x-3)(x+5)=0

=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)

=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)

=>\(3x-9+10x+5=90\)

=>13x-4=90

=>13x=94

=>\(x=\dfrac{94}{13}\)

d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))

=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)

=>3x-11=2x-4-x-1

=>3x-11=x-5

=>2x=6

=>x=3(nhận)

Câu 2/

Điều kiện xác định b tự làm nhé:

\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)

\(\Leftrightarrow x^4-25x^2+150=0\)

\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)

Tới đây b làm tiếp nhé.

a. ĐK: \(\frac{2x-1}{y+2}\ge0\)

Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)

\(\)Dấu bằng xảy ra khi  \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\) 

Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)

b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)

\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)

\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)

BALABOLO

TK NHA

a) Đ

b) S

c) S

d) Đ

\(x^2-2x+1< 9\)

\(\Leftrightarrow\left(x-1\right)^2< 9\)

\(\Leftrightarrow x-1< 3\)

\(\Leftrightarrow x< 4\)

\(\left(x-1\right)\left(4-x^2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(2-x\right)\left(2+x\right)\ge0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

\(\dfrac{x+2}{x-5}< 0\)

\(\Leftrightarrow x+2< 0\)

\(\Leftrightarrow x< -2\)

a)\(x^2-2x+1< 9\)

\(\Leftrightarrow\left(x-1\right)^2< 9\)

\(\Leftrightarrow\left(x-1\right)^2-9< 0\)

\(\Leftrightarrow\left(x-1-3\right)\left(x-1+3\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4< 0\\x+2>0\end{matrix}\right.hay\left[{}\begin{matrix}x-4>0\\x+2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 4\\x>-2\end{matrix}\right.hay\left[{}\begin{matrix}x>4\\x< -2\end{matrix}\right.\)(vô lý)

-Vậy nghiệm của BĐT là \(-2< x< 4\).

b) \(\left(x-1\right)\left(4-x^2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(2-x\right)\left(x+2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x-2>0\\x+2>0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1>0\\x-2< 0\\x+2>0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1>0\\x-2 >0\\x+2< 0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1< 0\\x-2< 0\\x+2< 0\end{matrix}\right.\)

 \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>2\\x>-2\end{matrix}\right.\) (vô lí) hay \(\left[{}\begin{matrix}x>1\\x< 2\\x>-2\end{matrix}\right.\) (có thể xảy ra) hay

\(\left[{}\begin{matrix}x>1\\x>2\\x< -2\end{matrix}\right.\) (vô lí) hay \(\left[{}\begin{matrix}x< 1\\x< 2\\x< -2\end{matrix}\right.\) (có thể xảy ra)

-Vậy nghiệm của BĐT là \(x< -2\) hay \(1< x< 2\).

c) ĐKXĐ: \(x\ne5\)

 \(\dfrac{x+2}{x-5}< 0\Leftrightarrow\left[{}\begin{matrix}x+2< 0\\x-5>0\end{matrix}\right.hay\left[{}\begin{matrix}x+2>0\\x-5< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -2\\x>5\end{matrix}\right.\)(vô lí) hay

\(\left[{}\begin{matrix}x>-2\\x< 5\end{matrix}\right.\) (có thể xảy ra)

-Vậy nghiệm của BĐT là \(-2< x< 5\)

\(\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

⇒ \(3x=-5\)

⇒ \(x=-\dfrac{5}{3}\)

\(a,\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}3x=-5\\x+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\-\dfrac{5}{3}+2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\2y=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}3x+5y=11\\2x+5y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=11\\3x+5y-2x-5y=11-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x=2\end{matrix}\right.\)

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

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)

=>x(2x+10-x+2)=0

=>x(x+12)=0

=>x=0 hoặc x=-12

b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

=>(2x-5)(3x+12)=0

=>x=5/2 hoặc x=-4

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

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

=>x=3 hoặc x=-1

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

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

=>(x+2)(-5x+3)=0

=>x=-2 hoặc x=3/5

\(a,\left(x-2\right)x=2x\left(x+5\right)\)

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

\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)