1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

a, \(x^2\)  - 19 = 5.9

     \(x^2\) - 19 = 45

     \(x^2\)         = 45 + 19

     \(x^2\)         = 64

      \(x^2\)        = 8²

      \(x\)         = 8 

b, (2\(x\) + 1)³ = -0,001

    (2\(x\) + 1)³ = (-0,1)³

     2\(x\) + 1   = -0,1

     2\(x\)        = -0,1 - 1

     2\(x\)       = - 1,1

       \(x\)      = -1,1: 2

       \(x\)      = -  0,55

\(x^2-19=5\cdot9\\\Rightarrow x^2-19=45\\\Rightarrow x^2=45+19\\\Rightarrow x^2=64\\\Rightarrow x^2=(\pm8)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(---\)

\((2x+1)^3=-0,001\\\Rightarrow (2x+1)^3=(-0,1)^3\\\Rightarrow2x+1=-0,1\\\Rightarrow2x=-0,1-1\\\Rightarrow2x=-1,1\\\Rightarrow x=-1,1:2\\\Rightarrow x=\dfrac{-11}{20}\\---\)

\(\bigg(\dfrac56\bigg)^{2x-1}=\bigg(\dfrac56\bigg)^5\\\Rightarrow 2x-1=5\\\Rightarrow2x=5+1\\\Rightarrow2x=6\\\Rightarrow x=6:2\\\Rightarrow x=3\\---\)

\(\bigg(\dfrac13x-\dfrac23\bigg)^3=27\\\Rightarrow\bigg(\dfrac13x-\dfrac23\bigg)^3=3^3\\\Rightarrow\dfrac13x-\dfrac23=3\\\Rightarrow\dfrac13x=3+\dfrac23\\\Rightarrow\dfrac13x=\dfrac{11}{3}\\\Rightarrow x=\dfrac{11}{3}:\dfrac13\\\Rightarrow x=11\\---\)

\(\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac12\bigg)^{15}\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg[\bigg(\dfrac{1}{2}\bigg)^5\bigg]^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1^5}{2^5}\bigg)^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1}{32}\bigg)^3\\\Rightarrow x=3\\Toru\)

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

1) Ta có: \(\left(3-x^2\right)+6-2x=0\)

\(\Leftrightarrow3-x^2+6-2x=0\)

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

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

\(\Leftrightarrow x^2+2x+1=10\)

\(\Leftrightarrow\left(x+1\right)^2=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)

2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)

\(\Leftrightarrow10x-5+7=8-4x+2\)

\(\Leftrightarrow10x+4x=8+2+5-7\)

\(\Leftrightarrow14x=8\)

\(\Leftrightarrow x=\dfrac{4}{7}\)

Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)

3) Ta có: \(x^2-6x+4\left(x-6\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy: S={6;-4}

1. (3x - 5)² - (3x + 1)² = 8

=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8

=> -6(6x - 4) = 8

=> 6x - 4 = \(\dfrac{-4}{3}\)

\(\Rightarrow x=\dfrac{4}{9}\)

2) 2x(8x - 3) - (4x - 3)² = 27

=> 16x² - 6x - 16x² + 24x - 9 = 27

=> 18x - 9 = 27

=> x = 2

3) (2x - 3)² - (2x + 1)² = 3

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

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

=> 4x - 2 = \(\dfrac{-3}{4}\)

\(\Rightarrow x=\dfrac{5}{16}\)

4) (x + 5)² - x² = 45

=> (x + 5 - x)(x + 5 + x) = 45

=> 5(2x + 5) = 45

=> 2x + 5 = 9

=> x = 2

5) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 18

=> x³ - 9x² + 27x - 27 - x³ + 27 + 9(x² + 2x + 1) = 18

=> -9x² + 27x + 9x² + 18x + 9 = 18

=> 45x + 9 = 18

=> 45x = 9

=> x = \(\dfrac{1}{5}\)

6) x(x - 4)(x + 4) - (x - 5)(x² + 5x + 25) = 13

=> x (x² - 16) - (x³ - 125) = 13

=> x³ - 16x - x³ + 125 = 13

=> -16x = -112

=> x = 7.

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

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

<=>\(8x=-16\)

<=>\(x=-2\)

i. 

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>

<=>

<=>\(8x=-16\)

<=>x=-2

i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>

<=>10x=2

<=>

1: =>(x+3)(x-5)=0

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

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

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

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

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)