TH1: \(x\ge2\)

\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=4\)

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

\(\Leftrightarrow x^4-5x^2=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\sqrt{5}\left(loại\right)\\x=\sqrt{5}\end{matrix}\right.\)

TH2: \(x< 2\)

\(-\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)

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

\(\Leftrightarrow x^4-5x^2+8=0\)

\(\Leftrightarrow\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm)

Vậy \(x=\sqrt{5}\)

dell bt

Ta có :

\(\left(x-1\right)\left(x-12\right)=2\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow x^2-13x+12=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow x^2-13x+12=2x^2-10x+12\)

\(\Leftrightarrow x^2+2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)

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

Vậy : \(x\in\left\{0,-2\right\}\)

cách đoán mò ngây thơ nhưng KQ vẫn đúng == chỉ thiếu KQ thoi 

\(\left(x-1\right)\left(x-12\right)=2\left(x-2\right)\left(x-3\right)\)

\(x-x+12=2x-4x+12\)

\(x-x-2x+4x=12-12\)

\(2x=0\Leftrightarrow x=0\)

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

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

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)

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

\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)

\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)

bạn tải ảnh về r up lại đi bạn

\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)

\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)

\(\Leftrightarrow-28x+37\ge12\)

\(\Leftrightarrow-28x\ge12-37\)

\(\Leftrightarrow-28x\ge-25\)

\(\Leftrightarrow x\le\dfrac{25}{28}\)

Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)

b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)

\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)

\(\Leftrightarrow-6x\ge30\)

\(\Leftrightarrow x\le-5\)

Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)

\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)

\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)

\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)

\(\Leftrightarrow-11x+37< 0\)

\(\Leftrightarrow-11x< -37\)

\(\Leftrightarrow x>\dfrac{37}{11}\)

vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)

`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}

7,3, -6

ĐKXĐ: \(x\ne7;x\ne2\)

BPT \(\Leftrightarrow f\left(x\right)=\dfrac{\left(6-2x\right)^3\left(x+6\right)}{\left(x-7\right)^3}\le0\)

Lập bảng xét dấu ta có:

Từ đây ta thấy \(-6\le x\le3\) hoặc \(x>7\) thỏa mãn bất phương trình ban đầu.

Vậy...

Đặt a = x² + x ta đc:

\(a^2+4a-12=0\)

\(\Rightarrow a^2-2a+6a-12=0\)

\(\Rightarrow a\left(a-2\right)+6\left(a-2\right)=0\)

\(\Rightarrow\left(a-2\right)\left(a+6\right)=0\)

=> a - 2 = 0 => a = 2

hoặc a + 6 = 0 => a = -6

+) Với a = 2 => x² + x = 2 => x² + x - 2 = 0 => (x - 1)(x + 2) = 0=> x = 1 hoặc x = -2

+) Với a = -6 => x² + x = -6 => x² + x + 6 = 0 

Vì x² + x + 6 > 0 => pt vô nghiệm

Vậy x = 1 ; x = -2