b) 3x⁴-3x³+9x³-9x²-24x²+24x-48x+48

=3x³(x-1)+9x²(x-1)-24x(x-1)-48(x-1)

=(x-1)(3x³+9x²-24x-48)

=3(x-1)(x³+3x²-8x-16)

Chú ý rằng |a| = b với b > 0 thì a = b hoặc a = - b.

a ) x ∈ − 5 8 ; 5 8 b ) x ∈ 1 12 ; 1 4

c ) x ∈ − 17 30 ; 9 10 d ) x ∈ − 19 4 ; 11 2

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

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

3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

a) x⁴ - 10x³ - 15x² + 20x + 4

= x⁴ + 2x³ - 12x³ - 24x² + 9x² + 18x + 2x + 4

= x³(x + 2) - 12x²(x + 2) + 9x(x + 2) + 2(x + 2)

= (x + 2)(x³ - 12x² + 9x + 2)

b)

2x⁴ - 5x³ - 27x² + 25x + 50

= 2x³(x - 2) - x²(x - 2) - 25x(x - 2) - 25(x - 2)

= (x - 2)(2x³ - x² - 25x - 25)

c)\(3x^4+6x^3-33x^2-24x+48\)

\(=3\left(x^4+2x^3-11x^2-8x+16\right)\)

\(=3\left(x^4-x^3-4x^2+3x^3-3x^2-12x-4x^2+4x+16\right)\)

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

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

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

\(=3\left[x\left(x-1\right)+4\left(x-1\right)\right]\left(x^2-x-4\right)\)

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

\(a,\Rightarrow1-6x+9x^2-9x-9x^2=-29\\ \Rightarrow-15x=-30\Rightarrow x=2\\ b,\Rightarrow8x^3-12x^2+6x-1-x^2+4x-4=4x-25x^2-6\\ \Rightarrow8x^3+12x^2+6x+1=0\\ \Rightarrow\left(2x+1\right)^3=0\\ \Rightarrow2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

a.

\(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow2\left(4-3x\right)=10+4\)

\(\Leftrightarrow2\left(4-3x\right)=14\)

\(\Leftrightarrow4-3x=7\)

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)

b.

\(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow3\left(-x+7\right)=-18+12=-6\)

\(\Leftrightarrow-x+7=-6:3=-2\)

\(\Leftrightarrow x=9\)

c.

\(-45:5.\left(-3-2x\right)=3\)

\(\Leftrightarrow-9.\left(-3-2x\right)=3\)

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

\(\Leftrightarrow2x=-\dfrac{8}{3}\)

\(\Leftrightarrow x=-\dfrac{4}{3}\notin Z\left(loại\right)\)

Câu này em ghi sai đề?

d.

\(3x-28=x+36\)

\(\Leftrightarrow2x=28+36\)

\(\Leftrightarrow2x=64\)

\(\Leftrightarrow x=32\)

e.

\(\left(-12\right)^2.x=56+10.13x\)

\(\Leftrightarrow144x=56+130x\)

\(\Leftrightarrow144x-130x=56\)

\(\Leftrightarrow14x=56\)

\(\Leftrightarrow x=4\)

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

=>(5x-7)(5x-3)=0

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

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

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

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

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

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

`<=> 2x(3x - 1) - 3x + 1 = 0`

`<=> 2x(3x - 1) - (3x - 1) = 0`

`<=> (2x - 1)(3x-1) = 0`

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

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

`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`

`@` `\text {Ans}`

`\downarrow`

`5,`

`-4x^2 + 1/9 = 0`

`<=> -4x^2 = 0 - 1/9`

`<=> -4x^2 = -1/9`

`<=> 4x^2 = 1/9`

`<=> x^2 = 1/9 \div 4`

`<=> x^2 = 1/36`

`<=> x^2 = (+-1/6)^2`

`<=> x = +-1/36`

Vậy, `S = {1/36; -1/36}`

`6,`

`(x-1)^3 = 8`

`<=> (x-1)^3 = 2^3`

`<=> x-1=2`

`<=> x = 2 + 1`

`<=> x = 3`

Vậy, `S = {3}`

`7,`

`(2x-1)^3 + 27 = 0`

`<=> (2x - 1)^3 = -27`

`<=> (2x-1)^3 = (-3)^3`

`<=> 2x - 1 = -3`

`<=> 2x = -3 + 1`

`<=> 2x = -2`

`<=> x = -1`

Vậy,` S = {-1}`

`8,`

`125 + 1/8(x-1)^3 = 0`

`<=> 1/8(x-1)^3 = - 125`

`<=> (x-1)^3 = -125 \div 1/8`

`<=> (x-1)^3 = -1000`

`<=> (x-1)^3 = (-10)^3`

`<=> x - 1 = - 10`

`<=> x = -10+1`

`<=> x = -9`

Vậy, `S = {-9}.`

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)