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NH
tìm x:
(x+3)3 - (x(3x+1)2 + (2x+1)4x2 - (2x+1) - 3x2=54
(x-3)3 - (x-3)(x2+3x+9)+ 6(x+1)2 + 3x2 = -33
1
HC
26 tháng 7 2017
1.
(x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) - 3x2 = 54
x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 + 1 - 3x2 = 54
9x3 + 6x2 + 27x - 9x3 - 6x2 - x
= 54 - 27 - 1
26x = 26
x = 1
H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
MK
19 tháng 8 2017
Tìm x:
a/ \(\left(x+3\right)^3-x\left(3x-1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)
<=> \(x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1-3x^2-54=0\)<=> \(12x^2+26x-26=0\)
<=> \(\left[\begin{array}{} x=\dfrac{-13+\sqrt{481}}{12}\\ x=\dfrac{-13-\sqrt{481}}{12} \end{array} \right.\)
b/ \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2+33=0\)
<=> 39x+39=0
<=> x=-1
Trả lời:
\(\left(x^2-3x+2\right)^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow\left[x^2-\left(3x-2\right)\right]^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow\left(x^2\right)^3-3.\left(x^2\right)^2.\left(3x-2\right)+3.x^2.\left(3x-2\right)^2-\left(3x-2\right)^3=x^6-\left(3x-2\right)^2\)
\(\Leftrightarrow x^6-3x^4.\left(3x-2\right)+3x^2.\left(3x-2\right)^2-\left(3x-2\right)^3=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow3x^4.\left(3x-2\right)-3x^2.\left(3x-2\right)^2=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x^2-x-2x+2\right)=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left[x.\left(x-1\right)-2.\left(x-1\right)\right]=0\)
\(\Leftrightarrow3x^2.\left(3x-2\right).\left(x-1\right).\left(x-2\right)=0\)
\(3x^2=0\Leftrightarrow x^2=0\Leftrightarrow x=0\)
\(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(x-1=0\Leftrightarrow x=1\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(x\in\left\{0,\frac{2}{3},1,2\right\}\)
\(\Leftrightarrow x^6-9x^5+33x^4-63x^3+62x^2-36x+8=x^6-\left(3x-2\right)^3\)
\(\Leftrightarrow x^6-9x^5+33x^4-63x^3+66x^2-36x+8=x^6-27x^3+54x^2-36x+8\)
\(\Leftrightarrow-9x^5+33x^4-36x^3+12x^2=0\)
\(\Leftrightarrow-3x^2\left(x-2\right)\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow x=\left\{0;2;1;\frac{2}{3}\right\}\)