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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)
- Với x+1=0 =>x=-1
- Với x2+8x+19 =>vô nghiệm
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
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6
=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = 9 => x = 0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0
=> 3(6x2 - 5x +1) - (18x2 - 29x + 3) = 0
=> (18x2 -15x + 1) -(18x2 - 29x +3) = 0
=> 18x2 - 15x +1 -18x2 + 29x - 3 = 0
=> 14x = 0 => x = 0
a)(x+2)(x+3)-(x-2)(x+5)=6
x(x+3)+2(x+3)-x(x+5)+2(x+5)=6
x2+3x+2x+6-x2-5x+2x+10=6
(x2-x2)+(3x+2x-5x+2x)+(10+6)=6
2x+16=6
2x=6-16
2x=-10
x=-10/2
x=-5. Vậy x=-5
b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6
6x2+27x+4x+18-6x2-x-12x-2=7
(6x2-6x2)+(27x+4x-x-12x)+(18-2)=7
18x+16=7
18x=7-16
x=-9/18=-1/2. Vậy x=-1/2
c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0
(9x-3)(2x-1)-(2x-3)(9x-1)=0
9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0
18x2-9x-6x+3-18x2+2x+27x-3=0
(18x2-18x2)+(27x+2x-6x-9x)+(3-3)=0
14x=0
x=0/14
x=0. Vậy x=0
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6 => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = -9 => x = -0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0
=> 3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
=> 18x2 - 15x + 3 - 18x2 + 29x -3 = 0
=> 14x = 0 => x = 0.
Tìm x.
a) 9x^2 – 6x – 3 = 0
b) x^3 + 9x^2 + 27x + 19 = 0
c) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3
a) \(9x^2-6x-3=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x^3+9x^2+27x+19=0\)
\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))
c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)
\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)
a) \(x^3+9x^2+27x+19=0\)
\(\Rightarrow x^3+x^2+8x^2+8x+19x+19=0\)
\(\Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\)
Mà \(x^2+8x+19=x^2+2.x.4+16+3=\left(x+4\right)^2+3\)
Vì \(\left(x+4\right)^2\ge0\) với mọi x
\(3>0\)
\(\Rightarrow\left(x+4\right)^2+3>0\) với mọi x
=> ( x + 4 )2 + 3 vô nghiệm
=> x + 1 = 0
=> x = -1
Vậy x = -1
b) \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)
\(\Rightarrow\left(2x\right)^3+3.\left(2x\right)^2+3.2x+1+x\left(x^2-2^2\right)-9x\left(x^2-4x+4\right)+57=0\)
\(\Rightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)
\(\Rightarrow48x^2-34x+58=0\)
\(\Rightarrow2\left(24x^2-17x+29\right)=0\)
\(\Rightarrow24x^2-17x+29=0\)
... Tới đây mình bí luôn rồi, sorry 
Câu a : \(x^3+9x^2+27x+19=0\)
\(\Leftrightarrow\left(x^3+9x^2+27x+27\right)-8=0\)
\(\Leftrightarrow\left(x+3\right)^3-2^3=0\)
\(\Leftrightarrow\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+2^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)
\(\Leftrightarrow x+1=0\) ( Vì : \(x^2+8x+19>0\))
\(\Leftrightarrow x=-1\)
Vậy \(x=-1\)
Câu b : \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)
\(\Leftrightarrow48x^2-34x+58=0\)
\(\Rightarrow PTVN\)
Vậy ko có giá trị của x
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
\(a,9x^2-6x-3=0\)
\(\Leftrightarrow9x^2-6x+1-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2=4\)
\(\Rightarrow3x-1=\pm2\)
\(\hept{\begin{cases}3x-1=2\Rightarrow x=1\\3x-1=-2\Rightarrow x=\frac{-1}{3}\end{cases}}\)
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 \(x^2+8x+19>0\)
=> \(x+1=0\)
<=> \(x=-1\)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)
<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)
<=> \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)
<=> \(x^3-25x-x^3+2x^2+4x-8=3\)
<=> \(2x^2-21x-8=3\)
<=> \(2x^2-21x-11=0\)
<=> \(2x^2-22x+x-11=0\)
<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)
<=> \(\left(2x+1\right)\left(x-11\right)=0\)
<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)