\(a,9x^2-6x-3=0\)

\(\Leftrightarrow9x^2-6x+1-4=0\)

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

\(\Rightarrow3x-1=\pm2\)

\(\hept{\begin{cases}3x-1=2\Rightarrow x=1\\3x-1=-2\Rightarrow x=\frac{-1}{3}\end{cases}}\)

Vậy \(x=1\) hoặc \(x=\frac{-1}{3}\)

\(b,x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^3+9x^2+27x+27-8=0\)

\(\Leftrightarrow\left(x+3\right)^3=8\)

\(\Rightarrow x+3=2\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

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

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=\frac{-11}{25}\)

Vậy \(x=\frac{-11}{25}\)

\(9x^2-6x-3=0\)

<=> \(\left(3x\right)^2-2.3x.1+1-4=0\)

<=> \(\left(3x-1\right)^2-2^2=0\)

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

<=> \(\hept{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

\(x^3+9x^2+27x+19\) \(=0\)

<=>\(x^3+x^2+8x^2+8x+19x+19=0\)

<=> \(x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

<=> \(\left(x^2+8x+19\right)\left(x+1\right)=0\)

mà \(x^2+8x+19>0\)

=> \(x+1=0\)

<=> \(x=-1\)

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

<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)

<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)

<=>  \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)

<=> \(x^3-25x-x^3+2x^2+4x-8=3\)

<=> \(2x^2-21x-8=3\)

<=> \(2x^2-21x-11=0\)

<=> \(2x^2-22x+x-11=0\)

<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)

<=> \(\left(2x+1\right)\left(x-11\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)

Bài 1:

a)-x^2+4x-5

=-(x²-4x+5)<0 với mọi x

=>-x^2+4x-5<0 với mọi x

b)x^4+3x^2+3

\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x

=>x^4+3x^2+3>0 với mọi x

c) bn xét từng th ra

Bài 2:

a)9x^2-6x-3=0

=>3(3x²-2x-1)=0

=>3x²-2x-1=0

=>3x²+x-3x-1=0

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

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

b)x^3+9x^2+27x+19=0

=>(x+1)(x²+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)

c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3

=>x³-25x-x³-8=3

=>-25x-8=3

=>-25x=1

=>x=-11/25

mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là

=>-25x=11

ghi rõ các bược hộ em ạ 

a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x

=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x

=-3

vậy...

a) Ta có: \(\left(2x+1\right)\left(4x-3\right)-6x\left(x+5\right)-2x\left(x-7\right)+18x\)

\(=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x\)

\(=-3\)

b) Ta có: \(9x\left(2x-5\right)-\left(6x+2\right)\left(3x-2\right)+39x\)

\(=18x^2-45x-18x^2+12x-6x+4+39x\)

\(=4\)

c) Ta có: \(4x\left(2x-3\right)+x\left(x+2\right)-9x\left(x-1\right)+x-5\)

\(=8x^2-12x+x^2+2x-9x^2+9x+x-5\)

\(=-5\)

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

b) x³ + 9x² + 27x + 19=0

x³ + 3. x² .1 + 3x . 3² + 3³ - 8 =0

(x + 3)³ - 2³ =0

(x+3)³ = 8

(x+3)³ = 2³

x+3 =2

x= -1

Vậy x  =- 1

b) x(x+5)(x-5) - (x+2)(x² - 2x +4) =3

x(x² - 25) - (x³ + 2³) =3

x³ - 25x - x³ -8 =3

-25x - 8 =3

-25x = 11

x= -11/25

Vậy x = -11/25

\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

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

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

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

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)