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2x3 + 3x2 + 6x + 5 = 02
<=> 2x3 + x2 + 5x + 2x2 + x + 5 = 0
<=> x(2x2 + x + 5) + (2x2 + x + 5) = 0
<=> (2x2 + x + 5)(x + 1) = 0
<=> x + 1 = 0 (vì 2x2 + x + 5 \(\ge\) 4,875 > 0 \(\forall\) x)
<=> x = - 1
Vậy tập nghiệm của pt là \(S=\left\{-1\right\}\)
b) 4x4 + 12x3 + 5x2 - 6x - 15 = 0
<=> 4x4 + 10x3 + 2x3 + 5x2 - 6x - 15 = 0
<=> 2x3(2x + 5) + x2(2x + 5) - 3(2x + 5) = 0
<=> (2x + 5)(2x3 + x2 - 3) = 0
<=> (2x + 5)(2x3 - 2x2 + 3x2 - 3) = 0
<=> (2x + 5)(x - 1)(2x2 + 3x + 3) = 0
<=> (2x + 5)(x - 1)[x2 + (x + 3/2)2 + 3/4]= 0
Mà x2 + (x + 3/2)2 + 3/4 > 0\(\forall x\)
\(\Rightarrow\left[\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=1\end{matrix}\right.\)
Vậy ...
a)\(4x^4+y^4=\left(4x^4+y^4+4x^2y^2\right)-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)
b)\(\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)
Đặt x^2 - 3x - 1 = A
\(\Rightarrow A^2-12A+27=\left(A^2-12A+36\right)-9\)
\(=\left(A-6\right)^2-9=\left(A-6-3\right)\left(A-6+3\right)\)
\(=\left(A-9\right)\left(A-3\right)\)
Hay \(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)
c)\(x^3-x^2-5x+125\)
\(=\left(x^3+5^3\right)-\left(x^2+5x\right)\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
d)\(xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
Mình có việc bận nên chỉ đưa được kết quả ý d) thật lòng mong các bạn tự tham khảo và giải
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
a, \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3-x+2x^2-1\)
b, \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
yêu cầu làm gì bạn