`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`

`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{3x-2-6}{3}=\frac{4x+1}{4}\)

=>\(\frac{3x-8}{3}=\frac{4x+1}{4}\)

=>3(4x+1)=4(3x-8)

=>12x+3=12x-32

=>3=-32(vô lý)

=>Phương trình vô nghiệm

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

=>3(x-3)+4(2x-1)=2(2-x)

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34+\frac13x+\frac23=3\)

=>\(x\left(\frac12+\frac14+\frac13\right)+\frac{6}{12}+\frac{9}{12}+\frac{8}{12}=3\)

=>\(x\left(\frac{6}{12}+\frac{3}{12}+\frac{4}{12}\right)=3-\frac{23}{12}=\frac{36}{12}-\frac{23}{12}=\frac{13}{12}\)

=>\(x\cdot\frac{13}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{x+4}{5}+\frac{5\left(-x+4\right)}{5}=\frac{2x-3\left(x-2\right)}{6}\)

=>\(\frac{x+4-5x+20}{5}=\frac{2x-3x+6}{6}\)

=>\(\frac{-4x+24}{5}=\frac{-x+6}{6}\)

=>6(-4x+24)=5(-x+6)

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=\frac{2\left(-x+1\right)}{2}\)

=>\(\frac{4-5x}{6}=-x+1\)

=>6(-x+1)=-5x+4

=>-6x+6=-5x+4

=>-6x+5x=4-6

=>-x=-2

=>x=2

6: \(-\left(\frac{x-3}{2}-2\right)=\frac{5\left(x+2\right)}{4}\)

=>\(-\frac{x-3-4}{2}=\frac{5\left(x+2\right)}{4}\)

=>\(\frac{-2\left(x-7\right)}{4}=\frac{5\left(x+2\right)}{4}\)

=>5(x+2)=-2(x-7)

=>5x+10=-2x+14

=>7x=4

=>x=4/7

7: \(\frac{2\left(2x+1\right)}{5}-\frac{6+x}{3}=\frac{5-4x}{15}\)

=>\(\frac{6\left(2x+1\right)-5\left(x+6\right)}{15}=\frac{5-4x}{15}\)

=>6(2x+1)-5(x+6)=-4x+5

=>12x+6-5x-30=-4x+5

=>7x-24=-4x+5

=>7x+4x=5+24

=>11x=29

=>\(x=\frac{29}{11}\)

8: \(\frac{7-3x}{2}-\frac{5+x}{5}=1\)

=>\(\frac{5\left(7-3x\right)-2\left(x+5\right)}{10}=1\)

=>5(7-3x)-2(x+5)=10

=>35-15x-2x-10=10

=>-17x+25=10

=>-17x=-15

=>x=15/17

1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)

\(\Leftrightarrow25x^2-1=25x^2-7x+15\)

\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)

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

\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Leftrightarrow5x=0\Leftrightarrow x=0\)

a/ \(\left(x+2\right)^2-9=0\)

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

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

<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

b/ \(x^2-2x+1=25\)

<=> \(\left(x-1\right)^2=25\)

<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

a) (x+2)²=0

 ==> x+2=0

 ==> x=0-2

==> x=-2

   x(x+4)(4-x)+(x-5)(x²+5x+25)=3

   (x²+4x)(4-x)+x³+5x²+25x-5x²-25x-125=3

   4x²-x³+16x-4x²+x³+5x²+25x-5x²-25x=3+125

   (4x²-4x²)-(x³-x³)+(16x+25x-25x)+(5x²-5x²)=128

   16x=128

   x=128:16

   x=8

Ko thấy j hết á bạn

1)

\(3\left(x-2\right)+4\left(x-1\right)=25\)

\(3x-6+4x-4=25\)

\(7x-10=25\\ 7x=35\\ x=5\)

2)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)

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

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

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

\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

3)

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

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

\(x^2-4x+4=4x^2-8x+4\)

\(x^2-4x+4-4x^2+8x-4=0\)

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

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

\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

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

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

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\) 

\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\) 

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

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

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

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