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(x+1) * (x2 +x+1) * (x-1) * (x2-x+1) = 7
[(x+1) * (x2 +x+1) ]*[(x-1) * (x2-x+1)]= 7 [Áp dụng hằng đẳng thức a3+b3=(a+b)*(a2+ab+b2)]
(x3+13) * (x3-13) = 7
x3 * x3 - x3 * 13 + x3 * 13 - 13 *13 =7
(x3)2 - 1 = 7
(x3)2 =7+1
(x3)2 =8
suy ra x = 3 căn 2
Lời giải:
$3x(1-x)+(x+3)(x-2)=-2(x-4)^2$
$\Leftrightarrow (3x-3x^2)+(x^2-2x+3x-6)=-2(x^2-8x+16)$
$\Leftrightarrow -2x^2+4x-6=-2x^2+16x-32$
$\Leftrightarrow 12x=26\Rightarrow x=\frac{13}{6}$
Vậy........
Ta có : \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x-4\right)^2\)
=> \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x^2-8x+16\right)\)
=> \(3x-3x^2+x^2+3x-2x-6=-2x^2+16x-32\)
=> \(3x-3x^2+x^2+3x-2x-6+2x^2-16x+32=0\)
=> \(-12x+26=0\)
=> \(x=\frac{26}{12}=\frac{13}{6}\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{13}{6}\right\}\)
a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)
\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{x+1}{x}\right)+\frac{1}{\left(x+1\right)^2}.\left(\frac{x^2+1}{x^2}\right)\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{2}{\left(x+1\right)^2x}+\frac{x^2+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{x^3}\)
\(P=\left[\frac{x^2+2x+1}{\left[x\left(x+1\right)\right]^2}\right]:\frac{x-1}{3}\)
\(P=\frac{\left(x+1\right)^2}{x^2\left(x+1\right)^2}:\frac{x-1}{3}\)
\(P=\frac{3}{x^2\left(x-1\right)}\)
b)Bài này liên quan đến dấu lớn nên mk ko làm đc
\(\Leftrightarrow\frac{4x^2}{5}\times\frac{2x-3}{6}-\frac{3x-10}{15}\times\frac{4x^2+3}{3}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{4x^2\left(2x-3\right)}{30}-\frac{\left(3x-10\right)\left(4x^2+3\right)}{45}=\frac{22x^2}{45}\)
\(\Leftrightarrow\frac{12x^2\left(2x-3\right)}{90}-\frac{2\left(3x-10\right)\left(4x^2+3\right)}{90}=\frac{44x^2}{90}\)
\(\Leftrightarrow12x^2\left(2x-3\right)-2\left(3x-10\right)\left(4x^2+3\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-2\left(12x^3+9x-40x^2-30\right)=44x^2\)
\(\Leftrightarrow24x^2-36x^2-24x^3-18x+80x^2+60=44x^2\)
\(\Leftrightarrow24x^3-36x^2-24x^3-18x+80x^2-44x^2=-60\)
\(\Leftrightarrow\left(24x^3-24x^3\right)+\left(-36x^2+80x^2-44x^2\right)-18x=-60\)
\(\Leftrightarrow-18x=-60\)
\(\Leftrightarrow x=\frac{-60}{-18}\)
\(\Leftrightarrow x=\frac{10}{3}\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(6\left(x+1\right)^2-2\left(x+1\right)^3+2\left(x-1\right)\left(x^2+x+1\right)=1\)
⇔ \(6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)+2\left(x^3-1\right)=1\)
⇔ \(6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\)
⇔ 6x + 1 = 0
⇔ x = \(\dfrac{-1}{6}\)
KL.........