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a)(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1-3x2)=54
\(\Rightarrow\)x3+9x2+27x+27-x(9x2+6x+1)+(2x+1)(x2-2x+1)=54
\(\Rightarrow\)x3+9x2+27x+27-9x3-6x2-x+2x3-4x2+2x+x2-2x+1=54
\(\Rightarrow\)-6x3+26x+28=54
\(\Rightarrow\)-6x3+26x=54-28
\(\Rightarrow\)-6x3+26x=26
\(\Rightarrow\)-6x3+26x-26=0
\(\Rightarrow\)-2(3x3+13x+14)
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow6\left(x+1\right)^2+3x^2=-33\text{ vô lý }\left(\text{vì }6\left(x+1\right)^2\ge0;3x^2\ge0\right)\)
\(\text{Vậy không có x nào thỏa mãn}\)
a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
hay x=7
b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)
\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)
=>39x+6=15
hay x=3/13
c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
hay x=14
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
<=>(x3-9x2+27x-27)-(x3-33+6(x2+2x+1)=15
<=>x3-9x2+27x-27-x3+27+6x2+12x+6=15
<=>-3x2+39x+9=0
<=>x2-13x+3=0
<=>(x2-2.x.13/2+169/4)-157/4=0
<=>(x-13/2)2=157/4
<=>x-13/2=\(\sqrt{\frac{157}{2}}\)hoặc x=13/2= - \(\sqrt{\frac{157}{2}}\)
<=>x=(13+\(\sqrt{\frac{157}{2}}\))hoặc x=\(\frac{13-\sqrt{\frac{157}{2}}}{2}\)
(x-3)3 - (x-3)(x2+3x+9) + 6(x+1)2 = 15
x3 -9x2 + 27x - 27 - (x3-27) + 6( x2+ 2x + 1) =15
x3 -9x2 + 27x - 27 - x3+ 27 + 6x2 + 12x + 6 = 15
-3x2 + 39x -9 = 0
-3(x2 - 13x + 3) = 0
x2 - 13x + 3 = 0
=> x=0,2350179569 ( chỗ này bấm máy tính)
còn giải thì làm theo mấy cách trong đây
BÀI 3 – 4
Phương trình bậc hai một ẩn – Công thức nghiệm
a) (3x+2)(2x+9) - (x+2)(6x+1) = (x+1) - (x-6)
<=> 6x2 + 27x + 4x + 18 - 6x2 - x - 12x - 2 = x+1 - x+6
<=> 18x + 16 = 7
<=> 18x = -9
<=> x = \(-\dfrac{1}{2}\)
b) 3(2x-1)(3x-1) - (2x-3)(9x-1) = 0
<=> 3.(6x2-2x-3x+1) - (18x2-2x-27x+3) = 0
<=> 3.(6x2-5x+1) - 18x2+29x-3 = 0
<=> 18x2-15x+3 - 18x2+29x - 3 = 0
<=> 14x = 0
<=> x = 0
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
<=> \(x^3-9x^2+27x-27\) \(-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)+3x^2=-33\)
<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
<=> \(-6x^2+39x+6=-33\)
<=> \(6x^2-39x-6=33\)
<=> \(6x^2-39x-39=0\)
<=> \(6\left(x^2-\frac{39}{6}x-\frac{39}{6}\right)=0\)
<=> \(x^2-2.x.\frac{39}{12}+\frac{1521}{144}-\frac{273}{16}=0\)
<=> \(\left(x-\frac{39}{12}\right)^2-\frac{273}{16}=0\)
<=> \(\left(x-\frac{39}{12}-\frac{\sqrt{273}}{4}\right)\left(x-\frac{39}{12}+\frac{\sqrt{273}}{4}\right)=0\)
<=> \(\left(x-\frac{13+\sqrt{273}}{4}\right).\left(x-\frac{13-\sqrt{273}}{4}\right)=0\)
<=> \(x=\frac{13+\sqrt{273}}{4}\) ( h ) \(x=\frac{13-\sqrt{273}}{4}\)
học tốt