`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) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)

\(=x^3+125-x^3+8x^2-16x+16x\)

\(=8x^2+125\)

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

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

a)

<=> 10x - 35 + 16x - 10 = 5 

<=> 10x + 16x = 5 + 35 + 10

<=> 26x = 50

<=> x = 50/26 = 25/13

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

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

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

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

a) 5.2² + (x + 3) = 5²

5.4 + x + 3 = 25

20 + x + 3 = 25

x + 23 = 25

x = 25 - 23

x = 2

b) 2³ + (x - 3²) = 5³ - 4³

8 + (x - 9) = 125 - 64

8 + x - 9 = 61

x - 1 = 61

x = 61 + 1

x = 62

c) 4.(x - 5) - 2³ = 2⁴.3

4x - 20 - 8 = 16.3

4x - 28 = 48

4x = 48 + 28

4x = 76

x = 76 : 4

x = 19

d) 5.(x + 7) - 10 = 2³.5

5x + 35 - 10 = 8.5

5x + 25 = 40

5x = 40 - 25

5x = 15

x = 15 : 5

x = 3

e) 7² - 7.(13 - x) = 14

49 - 91 + 7x = 14

7x - 42 = 14

7x = 14 + 42

7x = 56

x = 56 : 7

x = 8

a) \(5\cdot2^2+\left(x+3\right)=5^2\)

\(\Rightarrow x+3=5^2-5\cdot2^2\)

\(\Rightarrow x+3=25-5\cdot4\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

b) \(2^3+\left(x-3^2\right)=5^3-4^3\)

\(\Rightarrow8+\left(x-9\right)=125-64\)

\(\Rightarrow8+x-9=61\)

\(\Rightarrow x-1=61\)

\(\Rightarrow x=61+1\)

\(\Rightarrow x=62\)

c) \(4\left(x-5\right)-2^3=2^4\cdot3\)

\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)

\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)

\(\Rightarrow4\cdot\left(x-5\right)=56\)

\(\Rightarrow x-5=56:4\)

\(\Rightarrow x-5=14\)

\(\Rightarrow x=19\)

d) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow5\left(x+7\right)=50\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

e) \(7^2-7\left(13-x\right)=14\)

\(\Rightarrow7\left(13-x\right)=7^2-14\)

\(\Rightarrow7\left(13-x\right)=49-14\)

\(\Rightarrow7\left(13-x\right)=35\)

\(\Rightarrow13-x=5\)

\(\Rightarrow x=13-5\)

\(\Rightarrow x=8\)

f) \(5x-5^2=10\)

\(\Rightarrow5x=10+5^2\)

\(\Rightarrow5x=10+25\)

\(\Rightarrow5x=35\)

\(\Rightarrow x=\dfrac{35}{5}\)

\(\Rightarrow x=7\)

g) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x=3^4+2\cdot3^2\)

\(\Rightarrow9x=81+2\cdot9\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

h) \(10x+2^2\cdot5=10^2\)

\(\Rightarrow10x=10^2-2^2\cdot5\)

\(\Rightarrow10x=100-4\cdot5\)

\(\Rightarrow10x=80\)

\(\Rightarrow x=\dfrac{80}{10}\)

\(\Rightarrow x=8\)

i) \(125-5\left(4+x\right)=15\)

\(\Rightarrow5\left(4+x\right)=125-5\)

\(\Rightarrow5\left(4+x\right)=120\)

\(\Rightarrow4+x=\dfrac{120}{5}\)

\(\Rightarrow4+x=24\)

\(\Rightarrow x=24-4\)

\(\Rightarrow x=20\)

j) \(2^6+\left(5+x\right)=3^4\)

\(\Rightarrow5+x=3^4-2^6\)

\(\Rightarrow5+x=81-64\)

\(\Rightarrow5+x=17\)

\(\Rightarrow x=17-5\)

\(\Rightarrow x=12\)

f) 5x - 5² = 10

5x - 25 = 10

5x = 10 + 25

5x = 35

x = 35 : 5

x = 7

g) 9x - 2.3² = 3⁴

9x - 2.9 = 81

9x - 18 = 81

9x = 81 + 18

9x = 99

x = 99 : 9

x = 11

h) 10x - 2².5 = 10²

10x - 4.5 = 100

10x - 20 = 100

10x = 100 + 20

10x = 120

x = 120 : 10

x = 12

i) 125 - 5.(4 + x) = 15

5.(4 + x) = 125 - 15

5.(4 + x) = 110

4 + x = 110 : 5

4 + x = 22

X = 22 - 4

x = 18

j) 2⁶ + (5 + x) = 3⁴

64 + 5 + x = 81

69 + x = 81

x = 81 - 69

x = 12

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)