x^5 -x^3 -x^2 +1=0

x^3(x^2 -1 )-(x^2-1)=0

(x-1)(x^2+x+1)(x-1)(x+1)=0

(x-1)^2(x+1)(x^2+x+1)=0

=> x=1;x=-1

x^3- 5x^2+ 8x- 4= x^3- x^2- 4x^2+ 4x+ 4x- 4

                       = x^2(x-1)- 4x(x-1)+4(x-1)

                       = (x-1)(x^2-4x+4)

                       = (x-1)(x-1)^2

                       =(x-1)^3

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

Chúng ta sẽ giải từng phương trình một:

a. Đặt �=�2−2y=x2−2, ta có: (−2+�2)(�2−2)4=1(−2+x2)(

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

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

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

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

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

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

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

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

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

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a)    (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x² + 3x + 2x + 6 - (x² + 5x - 2x - 10) = 0
<=> x² + 3x + 2x + 6 - x² - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8

b)    (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 - (3x² - 12x - 5x + 20) = 0
<=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 - 3x² + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=>   x = 22/5

c)    (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x² - 10x + (4x - 8)(x + 1) + 2(x² - 4) = 0
<=> 8x + 16 - 5x² - 10x + 4x² + 4x - 8x - 8 + 2x² - 8 = 0
<=> x² - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay     x - 6 = 0
                  I<=> x      = 6

d)    (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x² + 16x - 9x - 6 - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33
<=> 24x² + 16x - 9x - 6 - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0
<=> 10x² - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay      10x - 19 = 0
                  I <=> 10x       = 19
                  I <=>    x       = 19/10

a) \(\left|2x+1\right|=\left|1-x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=1-x\\2x+1=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)

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

\(\Rightarrow\orbr{\begin{cases}4x-3=5\\4x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=8\\4x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)

e) \(\left|\frac{5}{4}-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\Leftrightarrow\left|\frac{5}{8}x+\frac{3}{5}\right|=\frac{9}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{3}{5}=\frac{9}{4}\\\frac{5}{8}x+\frac{3}{5}=-\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{33}{20}\\\frac{5}{8}x=-\frac{57}{20}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{66}{25}\\x=-\frac{114}{25}\end{cases}}\)

\(\left|2x+1\right|=\left|1-x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=-x+1\\2x+1=x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+1\\2x-x=-1-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

b. \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=4+2\\5x+x=4-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

c. \(\left|2x-3\right|-\left|3x+2\right|=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=3+2\\2x+3x=3-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=5\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)

d, e tương tự

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3