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

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

\(\Leftrightarrow\left|x+1\right|+\left|\left(x+1\right)\left(x+3\right)\right|-x^3-1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

Vậy nghiệm phương trình là: {-1; -3}

=4^2.3-4y^2-4y

=4.(12-y^2-y)

hol tot 

nho k nhe ae

good luck

x.x^2.x^3=64

x^1.x^2.x^3=2^6

x^6=2^6

x=2

x^1+2+3=64=2^6

=>x=6

\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)

\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)

\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)

\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)

\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)

\(\Rightarrow-\frac{1}{12}\left(x-2018\right)=0\Leftrightarrow x=2018\)

               Bài làm :

Ta có :

\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)

\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)

\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)

\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)

\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)

\(\text{Vì : }\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\ne0\Rightarrow x-2018=0\)

\(\Rightarrow x=2018\)

Vậy x=2018

\(70,5-5,28:\frac{15}{2}\)

\(=70,5-5,28:7,5\)

\(=70,5-0,704\)

\(=69,796\)

705/10-528/10:15/2

=141/2-264/5:15/2

=141/2-164/25

=3525-328/50

=3197/50

\(48-4y^2-4y\)

\(=-\left(4y^2+4y-48\right)\)

\(=-\left[\left(2y+1\right)^2-49\right]\)

\(=-\left(2y+1-7\right)\left(2y+1+7\right)\)

\(=-\left(2y-6\right)\left(2y+8\right)\)

\(=-4\left(y-3\right)\left(y+4\right)\)

p>3 suy ra p có dạng là 6k =1 hoặc 6k +5 

nếu p  có dạng 6k+2 suy ra p+4 = 6k +2 +4=6k +6chia hết cho 6 là hợp số (loại)

suy ra p có dạng 6k+1

p+8 = 6k +1 +8 = 6k + 9 chia hết cho 3 

vậy p+8 là hợp số