`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}

\(\left|x-2\right|=\left|2x-3\right|\)

Nếu : \(\left\{{}\begin{matrix}2x-3\ge0\Leftrightarrow2x\ge3\Leftrightarrow x\ge\dfrac{3}{2}\\2x-3< 0\Leftrightarrow2x< 3\Leftrightarrow x< \dfrac{3}{2}\end{matrix}\right.\)

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

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

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

Vậy pt vô nghiệm

__

\(\left|5-x\right|=\left|x+2\right|\)

Nếu : \(\left\{{}\begin{matrix}x+2\ge0\Leftrightarrow x\ge-2\\x+2< 0\Leftrightarrow x< -2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5-x=x+2\\5-x=-\left(x+2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=2-5\\5-x=-x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\0=-7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(ktm\right)\\0=-7\left(ktm\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

x(x + 2)(x + 3)(x + 5) = 72

⇔ (x² + 5x)(x² + 5x + 6) - 72 = 0 (1)

Đặt u = x² + 5x

⇒ x² + 5x + 6 = u + 6

(1) ⇔ u.(u + 6) - 72 = 0

⇔ u² + 6u - 72 = 0

⇔ u² + 12u - 6u - 72 = 0

⇔ (u² + 12u) - (6u + 72) = 0

⇔ u(u + 12) - 6(u + 12) = 0

⇔ (u + 12)(u - 6) = 0

⇔ u + 12 = 0 hoặc u - 6 = 0

*) u + 12 = 0

⇔ u = -12

⇒ x² + 5x = -12

⇔ x² + 5x + 12 = 0

⇔ x² + 2.5x/2 + 25/4 + 23/4 = 0

⇔ (x + 5/2)² + 23/4 = 0 (vô lý)

*) u - 6 = 0

⇔ u = 6

⇒ x² + 5x = 6

⇔ x² + 5x - 6 = 0

⇔ x² - x + 6x - 6 = 0

⇔ (x² - x) + (6x - 6) = 0

⇔ x(x - 1) + 6(x - 1) = 0

⇔ (x - 1)(x + 6) = 0

⇔ x - 1 = 0 hoặc x + 6 = 0

**) x - 1 = 0

⇔ x = 1

**) x + 6 = 0

⇔ x = -6

Vậy S = {-6; 1}

a: =>x+3=x-2 hoặc x+3=2-x

=>2x=-1

=>x=-1/2

b: =>3x+7=x-2 hoặc 3x+7=-x+2

=>2x=-9 hoặc 4x=-5

=>x=-5/4 hoặc x=-9/2

c: =>|3x-4|=|2x-5|

=>3x-4=2x-5 hoặc 3x-4=-2x+5

=>x=-1 hoặc x=9/5

\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)

b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\) 

a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).