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a) 2( x - 1 )2 + ( x + 3 )2 = 3( x - 2 )( x + 1 )
<=> 2( x2 - 2x + 1 ) + x2 + 6x + 9 = 3( x2 - x - 2 )
<=> 2x2 - 4x + 2 + x2 + 6x + 9 = 3x2 - 3x - 6
<=> 2x2 - 4x + x2 + 6x - 3x2 + 3x = -6 - 2 - 9
<=> 5x = -17
<=> x = -17/5
b) ( x - 1 )2 - 2( x - 3 ) = ( x + 1 )2
<=> x2 - 2x + 1 - 2x + 6 = x2 + 2x + 1
<=> x2 - 2x - 2x - x2 - 2x = 1 - 1 - 6
<=> -6x = -6
<=> x = 1
c) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2 + 3x2 = -33
<=> x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 ) + 3x2 = -33
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6 + 3x2 = -33
<=> x3 - 9x2 + 27x - x3 + 6x2 + 12x + 3x2 = -33 - 27 + 27 - 6
<=> 39x = -39
<=> x = -1
a) Đặt \(a=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x+3=a+4\\x-2=a-1\\x+1=a+2\end{cases}}\)
Ta có: \(2a^2+\left(a+4\right)^2=3.\left(a-1\right)\left(a+2\right)\)
\(\Leftrightarrow2a^2+a^2+4a+4=3.\left(a^2+a-2\right)\)
\(\Leftrightarrow3a^2+4a+4=3a^2+3a-6\)
\(\Leftrightarrow a=-10\)
\(\Rightarrow x-1=-10\)
\(\Leftrightarrow x=-9\)
Vậy \(S=\left\{-9\right\}\)
b) Đặt \(b=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x-3=b-2\\x+1=b+2\end{cases}}\)
Ta có: \(b^2-2.\left(b-2\right)=\left(b+2\right)^2\)
\(\Leftrightarrow b^2-2b+4=b^2+4b+4\)
\(\Leftrightarrow-6b=0\)
\(\Leftrightarrow b=0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
c) Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x^2+2x+1\right)+3x^2+33=0\)
\(\Leftrightarrow6x^2+12x+6+3x^2+33=0\)
\(\Leftrightarrow9x^2+12x+39=0\)
\(\Leftrightarrow\left(9x^2+12x+4\right)+35=0\)
\(\Leftrightarrow\left(3x+2\right)^2+35=0\)
Vì \(\left(3x+2\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(3x+2\right)^2+35\ge35>0\forall x\)
mà \(\left(3x+2\right)^2+35=0\)
\(\Rightarrow\)\(\left(3x+2\right)^2+35=0\)vô nghiệm
Vậy \(S=\varnothing\)
1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
a) Ta có: \(\left(3x+5\right)^2-\left(x+3\right)^2-8x\left(x+3\right)=12\)
\(\Leftrightarrow9x^2+30x+25-x^2-6x-9-8x^2-24x-12=0\)
\(\Leftrightarrow4=0\) (vô lý)
=> pt vô nghiệm
b) \(\left(2x-5\right)^2-\left(x-2\right)^2-\left(x-1\right)\left(3x+2\right)=8\)
\(\Leftrightarrow4x^2-20x+25-x^2+4x-4-3x^2+x+2-8=0\)
\(\Leftrightarrow-15x=-13\)
\(\Rightarrow x=\frac{13}{15}\)
c) \(-2x\left(x+3\right)+\left(2x-5\right)^2=-3\left(x+2\right)\)
\(\Leftrightarrow-2x^2-6x+4x^2-20x+25+3x+6=0\)
\(\Leftrightarrow2x^2-23x+31=0\)
\(\Leftrightarrow2\left(x^2-\frac{23}{2}x+\frac{529}{16}\right)-\frac{281}{8}=0\)
\(\Leftrightarrow\left(x-\frac{23}{4}\right)^2-\left(\frac{\sqrt{281}}{4}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{23+\sqrt{281}}{4}\right)\left(x-\frac{23-\sqrt{281}}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{23+\sqrt{281}}{4}=0\\x-\frac{23-\sqrt{281}}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{23+\sqrt{281}}{4}\\x=\frac{23-\sqrt{281}}{4}\end{cases}}\)
a) (x - 2)3 + (3x - 1)(3x + 1) = (x + 1)3
<=> x3 - 6x2 + 12x - 8 + 9x2 - 1 = x3 + 3x2 + 3x + 1
<=> x3 + 3x2 + 12x - x3 - 3x2 - 3x = 1 + 9
<=> 9x = 10
<=> x = 10/9
vậy S = {10/9}
b) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> -5x2 + 2x - 10x + 5x2 + 11x = -22 + 1
<=> 3x = -21
<=> x = -7
Vậy S = {-7}
c) (x + 1)(2x - 3) = (2x - 1)(x + 5)
<=> 2x2 - x - 3 = 2x2 + 9x - 5
<=> 2x2 -x - 2x2 - 9x = -5 + 3
<=>-10x = -2
<=> x = 1/5 Vậy S = {1/5}
d) (x - 1) - (2x - 1) = 9 - x
<=> x - 1 - 2x + 1 = 9 - x
<=> -x + x = 9
<=> 0x = 9 (vô nghiệm)
=> pt vô nghiệm
e) (x - 3)(x + 4) - 2(3x - 2) = (x - 4)2
<=> x2 + x - 12 - 6x + 4 = x2 - 8x + 16
<=> x2 - 5x - x2 + 8x = 16 + 8
<=> 3x = 24
<=> x = 8
Vậy S = {8}
g) (x + 1)(x2 - x + 1) - 2x = x(x + 1)(x - 1)
<=> x3 + 1 - 2x = x3 - x
<=> x3 - 2x - x3 + x = -1
<=> -x = -1 <=> x = 1
Vậy S = {1}
a) ( x - 1 )2 + ( x - 2 )2 = 2( x + 4 )2 - ( 22x + 27 )
<=> x2 - 2x + 1 + x2 - 4x + 4 = 2( x2 + 8x + 16 ) - 22x - 27
<=> 2x2 - 6x + 5 = 2x2 + 16x + 32 - 22x - 27
<=> 2x2 - 6x - 2x2 - 16x + 22x = 32 - 27 - 5
<=> 0x = 0 ( đúng ∀ x ∈ R )
Vậy phương trình có vô số nghiệm
b) ( x + 2 )2 - 2( x - 3 ) = ( x + 1 )2
<=> x2 + 4x + 4 - 2x + 6 = x2 + 2x + 1
<=> x2 + 2x - x2 - 2x = 1 - 4 - 6
<=> 0x = -9 ( vô lí )
Vậy phương trình vô nghiệm
c) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
a)
\(x^2-2x+1+x^2-4x+4=2\left(x^2+8x+16\right)-22x-27\)
\(2x^2-6x+5=2x^2+16x+32-22x-27\)
\(-6x+5=-6x+5\)
\(0=0\left(llđ\forall x\right)\)
Vậy \(x=R\)
b)
\(x^2+4x+4-2x+6=x^2+2x+1\)
\(x^2+2x+10=x^2+2x+1\)
\(10=1\)
\(0=-9\left(sai\right)\)
Vậy phương trình vô nghiệm
c)
\(x^3+3x^2+3x+1-x^3-3x^2=2\)
\(3x+1=2\)
\(3x=1\)
\(x=\frac{1}{3}\)
a) ( x - 1 )3 - 4x( x + 1 )( x - 1 ) + 3( x - 1 )( x2 + x + 1 )
= x3 - 3x2 + 3x - 1 - 4x( x2 - 1 ) + 3( x3 - 13 )
= x3 - 3x2 + 3x - 1 - 4x3 + 4x + 3x3 - 3
= ( x3 - 4x3 + 3x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
b) ( x - 1 )( x - 2 )( 1 + x + x2 )( 4 + 2x + x2 )
= [ ( x - 1 )( 1 + x + x2 ) ][ ( x - 2 )( 4 + 2x + x2 ) ]
= [ ( x - 1 )( x2 + x + 1 ) ][ ( x - 2 )( x2 + 2x + 4 ) ]
= ( x3 - 13 )( x3 - 23 )
= ( x3 - 1 )( x3 - 8 )
= x6 - 9x3 + 8
c) ( x - 1 )3 + 3( x - 1 )( x2 + x + 1 ) - 4x( x + 1 )( x - 1 )
= x3 - 3x2 + 3x - 1 + 3( x3 - 13 ) - 4x( x2 - 1 )
= x3 - 3x2 + 3x - 1 + 3x3 - 3 - 4x3 + 4x
= ( x3 + 3x3 - 4x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
a,\(\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-3x^2+3x-1-4x\left(x^2-1\right)+3\left(x^3-1\right)\)
\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)
\(=-3x^2+7x-4\)
b,\(\left(x-1\right)\left(x-2\right)\left(1+x+x^2\right)\left(4+2x+x^2\right)\)
\(=\left[\left(x-1\right)\left(x^2+x+1\right)\right]\left[\left(x-2\right)\left(x^2+2x+4\right)\right]\)
\(=\left(x^3-1\right)\left(x^3-8\right)\)
\(=x^6-9x^3+8\)
c,\(\left(x-1\right)^3+3\left(x-1\right)\left(x^2+x+1\right)-4x\left(x+1\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1+3\left(x^3-1\right)-4\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1+3x^3-3-4x^3+4x\)
\(=-3x^2+7x-4\)
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
Bài 1:
a) (5x-4)(4x+6)=0
\(\Leftrightarrow\orbr{\begin{cases}5x-4=0\\4x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=4\\4x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{5}\\y=\frac{-3}{2}\end{cases}}}\)
b) (x-5)(3-2x)(3x+4)=0
<=> x-5=0 hoặc 3-2x=0 hoặc 3x+4=0
<=> x=5 hoặc x=\(\frac{3}{2}\)hoặc x=\(\frac{-4}{3}\)
c) (2x+1)(x2+2)=0
=> 2x+1=0 (vì x2+2>0)
=> x=\(\frac{-1}{2}\)
bài 1:
a) (5x - 4)(4x + 6) = 0
<=> 5x - 4 = 0 hoặc 4x + 6 = 0
<=> 5x = 0 + 4 hoặc 4x = 0 - 6
<=> 5x = 4 hoặc 4x = -6
<=> x = 4/5 hoặc x = -6/4 = -3/2
b) (x - 5)(3 - 2x)(3x + 4) = 0
<=> x - 5 = 0 hoặc 3 - 2x = 0 hoặc 3x + 4 = 0
<=> x = 0 + 5 hoặc -2x = 0 - 3 hoặc 3x = 0 - 4
<=> x = 5 hoặc -2x = -3 hoặc 3x = -4
<=> x = 5 hoặc x = 3/2 hoặc x = 4/3
c) (2x + 1)(x^2 + 2) = 0
vì x^2 + 2 > 0 nên:
<=> 2x + 1 = 0
<=> 2x = 0 - 1
<=> 2x = -1
<=> x = -1/2
bài 2:
a) (2x + 7)^2 = 9(x + 2)^2
<=> 4x^2 + 28x + 49 = 9x^2 + 36x + 36
<=> 4x^2 + 28x + 49 - 9x^2 - 36x - 36 = 0
<=> -5x^2 - 8x + 13 = 0
<=> (-5x - 13)(x - 1) = 0
<=> 5x + 13 = 0 hoặc x - 1 = 0
<=> 5x = 0 - 13 hoặc x = 0 + 1
<=> 5x = -13 hoặc x = 1
<=> x = -13/5 hoặc x = 1
b) (x^2 - 1)(x + 2)(x - 3) = (x - 1)(x^2 - 4)(x + 5)
<=> x^4 - x^3 - 7x^2 + x + 6 = x^4 + 4x^3 - 9x^2 - 16x + 20
<=> x^4 - x^3 - 7x^2 + x + 6 - x^4 - 4x^3 + 9x^2 + 16x - 20 = 0
<=> -5x^3 - 2x^2 + 17x - 14 = 0
<=> (-x + 1)(x + 2)(5x - 7) = 0
<=> x - 1 = 0 hoặc x + 2 = 0 hoặc 5x - 7 = 0
<=> x = 0 + 1 hoặc x = 0 - 2 hoặc 5x = 0 + 7
<=> x = 1 hoặc x = -2 hoặc 5x = 7
<=> x = 1 hoặc x = -2 hoặc x = 7/5
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) ( x + 3 )2 - ( x - 4 )( x + 8 ) = 1
<=> x2 + 6x + 9 - ( x2 + 4x - 32 ) = 1
<=> x2 + 6x + 9 - x2 - 4x + 32 = 1
<=> 2x + 41 = 1
<=> 2x = -40
<=> x = -20
b) 3( x + 2 )2 + ( 2x - 1 )2 - 7( x + 3 )( x - 3 ) = 36
<=> 3( x2 + 4x + 4 ) + 4x2 - 4x + 1 - 7( x2 - 9 ) = 36
<=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
<=> 8x + 76 = 36
<=> 8x = -40
<=> x = -5
c) ( x - 3 )( x2 + 3x + 9 ) + x( x + 2 )( 2 - x ) = 1
<=> x3 - 27 - x( x + 2 )( x - 2 ) = 1
<=> x3 - 27 - x( x2 - 4 ) = 1
<=> x3 - 27 - x3 + 4x = 1
<=> 4x - 27 = 1
<=> 4x = 28
<=> x = 7