\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

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

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)

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

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

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

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

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

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

a.

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)

\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)

\(18x-2=16\)

\(18x=16+2\)

\(18x=18\)

\(x=\frac{18}{18}\)

\(x=1\)

b.

\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3=8\)

\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)

\(-4x=8-3\)

\(-4x=5\)

\(x=-\frac{5}{4}\)

c.

\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)

\(42x=41\)

\(x=\frac{41}{42}\)

- mọi người ơi giúp mình với ạ. ai mình cx cho đúng

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

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

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

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

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

Câu 1 :

a, Ta có : \(x^2-10x=-25\)

=> \(x^2-10x+25=0\)

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

=> \(x-5=0\)

=> \(x=5\)

Vậy phương trình có nghiệm là x = 5 .

b, Ta có : \(5x\left(x-1\right)=x-1\)

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

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

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

=> \(\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)

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

Vậy phương trình có nghiệm là x = 1, x = \(\frac{1}{5}.\)

c, Ta có : \(2\left(x+5\right)-x^2-5x=0\)

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

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

=> \(\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\)

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

Vậy phương trình có nghiệm là x = 2, x = -5 .

d, Ta có : \(x^2-2x-3=0\)

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

=> \(x\left(x+1\right)-3\left(x+1\right)=0\)

=> \(\left(x-3\right)\left(x+1\right)=0\)

=> \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)

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

Vậy phương trình có nghiệm là x = 3, x = -1 .

e, Ta có : \(2x^2+5x-3=0\)

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

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

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

=> \(\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = -3, x = \(\frac{1}{2}.\)

\(1.x^2-10x=-25\\ \Leftrightarrow x^2-10x+25=0\\\Leftrightarrow \left(x-5\right)^2=0\\\Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)

Vậy nghiệm của phương trình trên là \(5\)

\(2.5x\left(x-1\right)=x-1\\ \Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=1\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{1;\frac{1}{5}\right\}\)

\(3.2\left(x+5\right)-x^2-5x=0\\\Leftrightarrow 2x+10-x^2-5x=0\\ \Leftrightarrow-x^2-3x+10=0\\\Leftrightarrow x^2+3x-10=0\\\Leftrightarrow x^2-2x+5x-10=0\\\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\\\Leftrightarrow \left(x+5\right)\left(x-2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-5;2\right\}\)

\(4.x^2-2x-3=0\\\Leftrightarrow x^2+x-3x-3=0\\\Leftrightarrow \left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;3\right\}\)

\(5.2x^2+5x-3=0\\ \Leftrightarrow2\left(x^2+\frac{5}{2}x-\frac{3}{2}\right)=0\\ \Leftrightarrow x^2+\frac{5}{2}x-\frac{3}{2}=0\\ \Leftrightarrow x^2-\frac{1}{2}x+3x-\frac{3}{2}=0\\\Leftrightarrow x\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{2}\right)=0\\\Leftrightarrow \left(x+3\right)\left(x-\frac{1}{2}\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+3=0\\x-\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-3;\frac{1}{2}\right\}\)

a)4(x+2)-7(2x-1)+9(3x-4)=30                                                     b)2(5x-8)-3(4x-5)=4(3x-4)+11

<=>4x+8-14x+7+27x-36=30                                            <=>10x-16-12x+15=12x-16+11

<=>17x-21=30                                                                    <=>  -14x=-4     <=>x=2/7

<=>17x=51

<=>x=3 

phần cvaf d đâu bạn ơi

phần c  hình như có vấn đề

d)(5x-3).4x-2x(10x-3)=15

<=>20x²-12x-20x²+6x=15

<=>   -6x=15

<=>x=-15/6

a) x² + 10x - 2x - 20 = 0

=> x(x + 10) - 2(x + 10) = 0

=> (x - 2)(x + 10) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+10=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)

b) \(x^2-5x-24=0\)

\(\Rightarrow x^2-5x+\frac{25}{4}-\frac{121}{4}=0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2=\frac{121}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{5}{2}\right)^2=\left(-\frac{11}{2}\right)^2\\\left(x-\frac{5}{2}\right)^2=\left(\frac{11}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\left(-\frac{11}{2}\right)\\x-\frac{5}{2}=\frac{11}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{6}{2}=3\\x=\frac{16}{2}=8\end{cases}}\)

c) x² - 8x + 3x - 24 = 0

=> x(x - 8) + 3(x - 8) = 0

=> (x + 3)(x - 8) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

2x+3/5x+2 = 4x+5/10x+2

<=> (2x+3)(10x+2)=(5x+2)(4x+5)

<=>2x(10x+2)+3(10x+2)= 5x(4x+5)+2(4x+5)

<=> 20x^2+4x+20x+6 = 20x^2+25x+9x+10

<=> 20x^2+4x+20x+6 - (20x^2+25x+9x+10)=0 => 20x^2+24x+6-(20x^2+34x+10)=0

                                                                  <=>   -10x-4=0

                                                                 <=>-10x=4

                                                                  <=> x= -4/10

2x+3/5x+2=4x+5/10x+2

=> (2x+3)(10x+2)=(5x+2)(4x+5)

=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )

=> 34x+6=33x+10

=> 34x-33x=-6+10

=> x=4

x⁴+5x²-9x +3

Tìm x?