\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)

TH1: x<1

(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0

=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(5x+55=0\)

=>x=-11(nhận)

TH2: 1<=x<2

Phương trình (1) sẽ trở thành:

\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(7x+53=0\)

=>\(x=-\dfrac{53}{7}\left(loại\right)\)

TH3: 2<=x<3

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)

=>\(11x+45=0\)

=>\(x=-\dfrac{45}{11}\left(loại\right)\)

TH4: 3<=x<4

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)

=>\(-3x+27=0\)

=>x=9(loại)

TH5: 4<=x<5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)

=>\(25x-5=0\)

=>x=1/5(loại)

TH6: x>=5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)

=>35x-55=0

=>x=55/35(loại)

a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023

mình cũng ko biết giải 

ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)

\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)

\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)

\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)

\(\Leftrightarrow x^3+10x^2-32x-384=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)

\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)

x=6

cần lời giải ko

Điều kiện: x+ 2 \(\ne\) 0 ; x+ 4 \(\ne\) 0; x+ 8 \(\ne\) 0 ; x + 14 \(\ne\) 0

<=> \(\frac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\frac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\frac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{x+4}{\left(x+2\right)\left(x+4\right)}-\frac{x+2}{\left(x+2\right)\left(x+4\right)}+\frac{x+8}{\left(x+4\right)\left(x+8\right)}-\frac{x+4}{\left(x+4\right)\left(x+8\right)}+\frac{x+14}{\left(x+8\right)\left(x+14\right)}-\frac{x+8}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)<=> \(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{12\left(x+4\right)}{\left(x+2\right)\left(x+14\right)\left(x+4\right)}=\frac{x\left(x+14\right)}{\left(x+2\right)\left(x+4\right)\left(x+14\right)}\)

<=> 12(x + 4) = x (x + 14)

<=> 12x + 48 = x² + 14 x

<=> x² + 2x - 48 = 0 

<=> x² + 8x - 6x - 48 = 0 

<=> x(x + 8) - 6 (x + 8) = 0 

<=> (x - 6)(x + 8) = 0 <=> x - 6 = 0 (do x + 8 \(\ne\) 0)

<=> x = 6 

Vậy x = 6

1: Ta có: \(4x^2-36=0\)

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

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

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

Ta thấy:\(\begin{cases}\left|x-1\right|\ge0\\\left|x-2\right|\ge0\\\left|x-3\right|\ge0\end{cases}\)

\(\Rightarrow\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\)

\(\Rightarrow4\left(x-4\right)\ge0\Rightarrow x-4\ge0\Rightarrow x\ge4\)

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

\(\Rightarrow3x-6=4x-16\)

\(\Rightarrow x=10\) (thỏa mãn)

Vậy \(x=10\)

         (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

<=.2x²-8x+3x-12+x² -2x-5x+10=3x²-12x-5x+20

<=>5x=22

<=>x=22/5

vay

bấm vào casio => X=5

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

=> \(\left(2x+3\right)\left(x-4\right)-\left(3x+5\right)\left(x-4\right)-\left(x-5\right)\left(x-2\right)=0\)

=> \(\left(x-4\right)\left(-x-2\right)-\left(x-5\right)\left(x-2\right)=0\)

=> \(-x^2-2x+4x+8-x^2+2x+5x-10=0\)

=> \(-2x^2+9x-2=0\)

=> \(-\left(x^2-2x+1\right)-\left(x^2-2x+1\right)+5x=0\)

=> \(-\left(x-1\right)^2-\left(x-1\right)^2+5x=0\)

=> \(-2\left(x-1\right)^2+5x=0\)

=> \(2\left(x-1\right)^2-5x=0\)

...................

để mik giải rồi ghi lại sau nha 

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

\(\left(x-4\right)^2-\left(x-4\right)^4=0\)

\(\left(x-4\right)^2\left[1-\left(x-4\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-4=0\\\left(x-4\right)^2=1-0=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0+4=4\\\left(x-4\right)^2=\left(-1\right)^2\text{ Hoặc }\left(x-4\right)^2=1^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x-4=-1\text{ Hoặc }x-4=1\end{cases}}\)                  \(\Rightarrow\orbr{\begin{cases}x=4\\x=-1+4=3\text{ Hoặc }x=1+4=5\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{4\text{ ; }3\text{ ; }5\right\}\)

Cảm ơn bạn Lily

Bài giải

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

\(\left(x-4\right)^2-\left(x-4\right)^4=0\)

\(\left(x-4\right)^2\left[1-\left(x-4\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-4=0\\\left(x-4\right)^2=1-0=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0+4=4\\\left(x-4\right)^2=\left(-1\right)^2\text{ Hoặc }\left(x-4\right)^2=1^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x-4=-1\text{ Hoặc }x-4=1\end{cases}}\)                  \(\Rightarrow\orbr{\begin{cases}x=4\\x=-1+4=3\text{ Hoặc }x=1+4=5\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{4\text{ ; }3\text{ ; }5\right\}\)