a: \(y=\left(5x-10\right)^4\)

=>\(y'=4\cdot\left(5x-10\right)'\cdot\left(5x-10\right)^3\)

\(=4\cdot5\cdot\left(5x-10\right)^3=20\left(5x-10\right)^3\)

Đặt y'>0

=>\(20\left(5x-10\right)^3>0\)

=>\(\left(5x-10\right)^3>0\)

=>5x-10>0

=>x>2

Đặt y'<0

=>\(20\left(5x-10\right)^3< 0\)

=>\(\left(5x-10\right)^3< 0\)

=>5x-10<0

=>x<2

Vậy: hàm số đồng biến trên \(\left(2;+\infty\right)\)

Hàm số nghịch biến trên \(\left(-\infty;2\right)\)

c: \(y=\left(x^3-1\right)^3\)

=>\(y'=3\left(x^3-1\right)'\cdot\left(x^3-1\right)^2\)

\(=9x^2\left(x^3-1\right)^2>=0\forall x\)

=>Hàm số luôn đồng biến trên R

d: \(y=\left(x^2-1\right)\left(x+2\right)\)

=>\(y'=\left(x^2-1\right)'\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)

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

\(=2x^2+4x+x^2-1=3x^2+4x-1\)

Đặt y'>0

=>\(3x^2+4x-1>0\)

=>\(\left[{}\begin{matrix}x< \dfrac{-2-\sqrt{7}}{3}\\x>\dfrac{-2+\sqrt{7}}{3}\end{matrix}\right.\)

Đặt y'<0

=>\(3x^2+4x-1< 0\)

=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)

Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)

Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)

b: \(y=\left(-x-1\right)\left(x+2\right)^4\)

=>\(y'=\left(-x-1\right)'\left(x+2\right)^4+\left(-x-1\right)\left[\left(x+2\right)^4\right]'\)

\(=-\left(x+2\right)^4+\left(-x-1\right)\cdot4\left(x+2\right)'\left(x+2\right)^3\)

\(=-\left(x+2\right)^4+4\left(x+2\right)^3\cdot\left(-x-1\right)\)

\(=-\left(x+2\right)^3\left[\left(x+2\right)+4\left(x+1\right)\right]\)

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

Đặt y'<0

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

=>(x+2)(5x+6)>0

TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x>-\dfrac{6}{5}\)

TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x< -2\)

Đặt y'>0

=>(x+2)(5x+6)<0

TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow-2< x< -\dfrac{6}{5}\)

TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy: HSĐB trên các khoảng \(\left(-\infty;-2\right);\left(-\dfrac{6}{5};+\infty\right)\)

HSNB trên khoảng \(\left(-2;-\dfrac{6}{5}\right)\)

a: \(y=\left(x^2-1\right)^2\)

=>\(y'=2\left(x^2-1\right)'\left(x^2-1\right)\)

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

Đặt y'>0

=>\(x\left(x^2-1\right)>0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1>0\end{matrix}\right.\)

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

=>\(x>1\)

TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< 0\)

Đặt y'<0

=>\(x\left(x^2-1\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x^2< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\-1< x< 1\end{matrix}\right.\)

=>0<x<1

TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1>0\end{matrix}\right.\)

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

=>x<-1

Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-1;0\right)\)

Hàm số nghịch biến trên các khoảng (0;1) và \(\left(-\infty;-1\right)\)

b: \(y=\left(3x+4\right)^3\)

=>\(y'=3\left(3x+4\right)'\left(3x+4\right)^2\)

\(\Leftrightarrow y'=9\left(3x+4\right)^2>=0\forall x\)

=>Hàm số luôn đồng biến trên R

c: \(y=\left(x+3\right)^2\left(x-1\right)\)

=>\(y=\left(x^2+6x+9\right)\left(x-1\right)\)

=>\(y'=\left(x^2+6x+9\right)'\left(x-1\right)+\left(x^2+6x+9\right)\left(x-1\right)'\)

=>\(y'=\left(2x+6\right)\left(x-1\right)+x^2+6x+9\)

=>\(y'=2x^2-2x+6x-6+x^2+6x+9\)

=>\(y'=3x^2-2x+3\)

\(\Leftrightarrow y'=3\left(x^2-\dfrac{2}{3}x+1\right)\)

=>\(y'=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{8}{9}\right)\)

=>\(y'=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{3}>=\dfrac{8}{3}>0\forall x\)

=>Hàm số luôn đồng biến trên R

d: \(y=\left(2x+2\right)\left(x^3-1\right)\)

=>\(y'=\left(2x+2\right)'\left(x^3-1\right)+\left(2x+2\right)\left(x^3-1\right)'\)

\(=2\left(x^3-1\right)+3x^2\left(2x+2\right)\)

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

\(=8x^3+6x^2-2\)

Đặt y'>0

=>\(8x^3+6x^2-2>0\)

=>\(x>0,46\)

Đặt y'<0

=>\(8x^3+6x^2-2< 0\)

=>\(x< 0,46\)

Vậy: Hàm số đồng biến trên khoảng tầm \(\left(0,46;+\infty\right)\)

Hàm số nghịch biến trên khoảng tầm \(\left(-\infty;0,46\right)\)

Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:

$|x-1|+|x-4|=|x-1|+|4-x|\geq |x-1+4-x|=3$

$|x-2|+|y-3|\geq 0$

$\Rightarrow |x-1|+|x-2|+|y-3|+|x-4|\geq 3$

Dấu "=" xảy ra khi:
\(\left\{\begin{matrix} (x-1)(4-x)\geq 0\\ x-2=0\\ y-3=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=3\end{matrix}\right.\)

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow-4x+3+5x+2=0\)

\(\Leftrightarrow x=-5\)

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 \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)