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a/ x.(x + 1)(x2 + x + 1) = 42
=> (x2 + x)(x2 + x + 1) = 42
Đặt a = x2 + x ta đc:
a.(a + 1) = 42
=> a2 + a - 42 = 0
=> (a - 6)(a + 7) = 0
=> a = 6 hoặc a = -7
Với a = 6 => x2 + x = 6 => x2 + x - 6 = 0 => (x - 2)(x + 3) = 0 => x = 2 hoặc x = -3
Với a = -7 => x2 + x = -7 => x2 + x + 7 = 0 , mà x2 + x + 7 > 0 => pt vô nghiệm
Vậy x = 2 , x = -3
b/ (3x - 1)2 - 5(2x + 1)2 + (6x - 3)(2x + 1) = (x - 1)2
=> 9x2 - 6x + 1 - 5.(4x2 + 4x + 1) + (12x2 - 3) = x2 - 2x + 1
=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 12x2 - 3 - x2 + 2x - 1 = 0
=> - 24x - 8 = 0
=> -24x = 8
=> x = -1/3
Vậy x = -1/3
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
Nhớ ghi dấu ngoặc tránh giải sai.
\(a.\) \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}\)
Ta có:
\(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
nên \(MTC:\) \(2\left(x-3\right)\left(x+3\right)\)
Do đó: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
1) \(A=\left[x^4-\left(x-1\right)^2\right]:\left(x^2+x-1\right)-x^2+x=\left[\left(x^2-x+1\right)\left(x^2+x-1\right)\right]:\left(x^2+x-1\right)-x^2+x=x^2-x+1-x^2+x=1\)
2) \(B=\dfrac{\left(x+1\right)\left(x+2\right)+4\left(x-2\right)+2-7x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4}{x^2-4}=1\)
a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)
\(=\dfrac{\left(x-1\right)^3\cdot\left(x^2-x+1\right)}{-2x\left(x^2-2x+2\right)}\)
b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)
\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3+3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(\frac{x+2}{x+1}=\frac{x}{x+1}+\frac{2}{x+1}\)
\(\frac{2x-3}{x-1}=\frac{2x}{x-1}+\frac{-3}{x-1}\)
\(\frac{x^2-3x+5}{x+1}=\frac{x^2}{x+1}+\frac{-3x+5}{x+1}\)
câu b nè
\(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}-\frac{x+3}{x^2-1}\)
=\(\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
=\(\frac{\left(3x^2+x+3x+1\right)-\left(x^2-2x+1\right)-\left(x^2-x-3+3x\right)}{\left(x-1\right)^2\left(x+1\right)}\)
=\(\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}=\frac{x^2+4x+3}{\left(x+1\right)\left(x-1^2\right)}\)
=\(\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}=\frac{x+3}{\left(x-1\right)^2}\)