Bạn tham khảo cách giải:

Bạn tham khảo cách giải: 

Bạn tham khảo cách giải: 

Sửa đề: (x-3)(x+5)-(x-2)(x+2)+(x-2)^2+(x+3)^2-2(x-1)(x+1)

\(=x^2+2x-15-x^2+4+x^2-4x+4+\left(x+3\right)^2-2\left(x^2-1\right)\)

\(=x^2-2x-7+x^2+6x+9-2x^2+2\)

=4x+4

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)

ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)

\(\Leftrightarrow3x^2+15x-378=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)

@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T

\(\left(2^x-8\right)^3+\left(4^x+13\right)^3=\left(4^x+2^x+5\right)^3\)

\(\Leftrightarrow\left(2^x-8+4^x+13\right)\left[\left(2^x-8\right)^2-\left(2^x-8\right)\left(4^x+13\right)+\left(4^x+13\right)^2\right]=\left(4^x+2^x+5\right)^3\)

Tự khai triển tính nốt ra, chắc là ra đấy

`#3107`

`(x - 3)^5 = 4(x - 3)^3`

`=> (x - 3)^5 - 4(x - 3)^3 = 0`

`=> (x - 3)^3 * [ (x - 3)^2 - 4] = 0`

`=>`\(\left[{}\begin{matrix}\left(x-3\right)^3=0\\\left(x-3\right)^2-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2=4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=\left(\pm2\right)^2\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=3\\x=5\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 3; 5}.`

Lời giải:

Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$

$\Rightarrow x=2018a; y=2019a; z=2020a$

$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$

Mặt khác:

$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$

Từ $(1); (2)$ ta có đpcm.

Lời giải:
Đặt $|x+2|=a$ với $a\geq 0$. Khi đó:

$A=\frac{3+2a}{1+a}=\frac{2(1+a)+1}{1+a}=2+\frac{1}{1+a}$

Vì $a\geq 0$ với mọi $x$ nên $1+a\geq 1$

$\Rightarrow A=2+\frac{1}{1+a}\leq 2+\frac{1}{1}=3$

Vậy $A_{\max}=3$. Giá trị này đạt tại $a=0\Leftrightarrow |x+2|=0\Leftrightarrow x=-2$

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

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

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

\(=\)\(\left(3x-2\right)\left(3x-6\right)\)

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

Chúc bạn học tốt ~