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

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) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....

b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........

c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......

d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......

\(a,3x^3+6x^2-4x=0\)

\(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x^2+6x-4=0\left(1\right)\end{cases}}\)

\(\Delta_{\left(1\right)}=36+4\cdot3\cdot4=84>0\)

\(\text{\Rightarrow pt có 2 nghiệm phân biệt}\)

\(x_1=\frac{-3+\sqrt{21}}{3};x_2=\frac{-3-\sqrt{21}}{3}\)

\(\text{Vậy phương trình đã cho bằng 0 khi x=0 hoặc x= }\frac{-3\pm\sqrt{21}}{3}\)

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

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

\(\left(x+1\right)\left(x+1-1\right)=0\)

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

\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....

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

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

\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)

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

\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........

\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)

\(\left(x+6\right)\left(x-7\right)=0\)

\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....

\(x^3-5x^2+x-5=0\)

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

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

\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........

\(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^3+10x\right)=0\)

\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............

nhớ chọn mk nha

Bài 3:

a) Ta có: \(x^3-16x=0\)

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

\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{0;4;-4\right\}\)

b) Ta có: \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)

\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)

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

Vậy: \(x\in\left\{0;2\right\}\)

c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)

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

\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)

d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

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

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

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

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

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

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

Vậy: \(x\in\left\{1;2\right\}\)

một lượt tối đa 2 câu làm vậy có thánh nào dmas beensg tới

a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt \(x^2+3x+1=t\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-24=0\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

TH1:t=5\(\Rightarrow x^2+3x+1=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

TH2:t=-5\(\Rightarrow x^2+3x+1=-5\)

\(\Leftrightarrow x^2+3x+6=0\)(vô nghiệm)

Vậy ...

b)\(\Leftrightarrow2\left(x^4-10x^2+9\right)=0\)

\(\Leftrightarrow x^4-9x^2-x^2+9=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(x^2-1\right)=0\)

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

(*)\(\dfrac{15x\left(x+5\right)}{20x^2\left(x+5\right)}=\dfrac{3}{4x}\)

(*)\(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)

(*)\(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}\)