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\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
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
c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)
\(=x^2+x-10\)
a) \(x^3+2\left(x-1\right)^2-2\left(x-1\right)\left(x+1\right)=x^3+x-4-\left(x-7\right)\).
\(\Leftrightarrow x^3+2\left(x^2-2x+1\right)-2\left(x^2-1\right)=x^3+x-4-x+7\)
\(\Leftrightarrow x^3+2x^2-4x+2-2x^2+2=x^3+3\)
\(\Leftrightarrow x^3-4x+4=x^3+3\)
\(\Leftrightarrow4x-1=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{4}\right\}\)
b) \(2\left(x-3\right)+1=2\left(x+1\right)-9\)
\(\Leftrightarrow2x-6+1=2x+2-9\)
\(\Leftrightarrow2x-5=2x-7\)
\(\Leftrightarrow2=0\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
c) \(3\left(x+1\right)\left(x-1\right)-5=3x^2+2\)
\(\Leftrightarrow3\left(x^2-1\right)-5=3x^2+2\)
\(\Leftrightarrow3x^2-3-5=3x^2+2\)
\(\Leftrightarrow3x^2-8=3x^2+2\)
\(\Leftrightarrow0=10\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
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 \(S=\varnothing\)