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\(a,x=2\Leftrightarrow A=3\cdot4-4\cdot2-1=12-8-1=3\\ b,B=x^3-1-2x+x^2-2+x-x^3=x^2-x-3\\ c,C=B-A=x^2-x-3-3x^2+3x+1=-2x^2-2x-2\\ C=-2\left(x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\right)=-2\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{2}\le-\dfrac{3}{2}\\ C_{max}=-\dfrac{3}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
Để phương trình có nghiệm thì :
\(\Delta_x=a^2-\left(2a^2+b^2-5\right)\ge0\)
\(\Leftrightarrow a^2+b^2\le5\)
\(\Leftrightarrow\left(a+b\right)^2\le5+2ab\)
\(\Leftrightarrow ab\ge\frac{\left(a+b\right)^2-5}{2}\)
Ta có :
\(P=\left(a+1\right)\left(b+1\right)=ab+a+b+1\)
\(\ge\frac{\left(a+b\right)^2-5}{2}+\left(a+b\right)+1=\frac{1}{2}\left(a+b+1\right)^2-2\ge-2\)
Dấu " = " xảy ra khi \(\hept{\begin{cases}a=-2\\b=1\end{cases}}\)
a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)
= x³ - 125 - x² + 4 + x³ + x² + 4x
= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)
= 2x³ + 4x - 121
b) Tại x = -2 ta có:
A = 2.(-2)³ + 4.(-2) - 121
= 2.(-8) - 8 - 121
= -16 - 129
= -145
c) x² - 1 = 0
x² = 1
x = -1; x = 1
*) Tại x = -1 ta có:
A = 2.(-1)³ + 4.(-1) - 121
= 2.(-1) - 4 - 121
= -2 - 125
= -127
*) Tại x = 1 ta có:
A = 2.1³ + 4.1 - 121
= 2.1 + 4 - 121
= 2 - 117
= -115
Bài 1.
Ta có: $x+y+z=0$
$\Rightarrow x+y=-z,\ y+z=-x,\ x+z=-y$.
Suy ra: $N=(x+y)(y+z)(x+z)$$=(-z)(-x)(-y)$$=-xyz$.
Mà: $xyz=2$.
Nên: $N=-2$.
Vậy: $N=-2$.
Bài 2.
$\frac{a}{b}=\frac{10}{3}\Rightarrow a=10k,\; b=3k$
$\frac{3a-2b}{a-3b}=\frac{3\cdot10k-2\cdot3k}{10k-3\cdot3k}$
$=\frac{30k-6k}{10k-9k}$
$=\frac{24k}{k}$
$=24$
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
(ax-2(a+2))[a(x-1)+2]+2\(\left(-a^4+4\right)\) +\(3x\cdot a^2\)
\(=\left(ax-2a-4\right)\left(ax-a+2\right)+2\left(-a^4+4\right)+3a^2\cdot x\)
\(=a^2x^2-a^2x+2a\cdot x-2a^2\cdot x+2a^2-4a-4\cdot ax+4a-8\) +\(2\left(-a^4+4\right)+3a^2\cdot x\)
\(=a^2x^2-3a^2\cdot x-2a\cdot x+2a^2-8-2a^4+8+3a^2\cdot x\)
\(=a^2x^2-2a\cdot x-2a^4+2a^2\)
\(P=\frac{(ax-2(a+2))[a(x-1)+2]+2\left(-a^4+4\right)+3x\cdot a^2}{-2ax}\)
\(=\frac{a^2x^2-2a\cdot x-2a^4+2a^2}{-2ax}\)
Khi a=2; x=1 thì \(P=\frac{2^2\cdot1^2-2\cdot2\cdot1-2\cdot2^4+2\cdot2^2}{-2\cdot2\cdot1}=\frac{-2^5+2^3}{-4}=\frac{-32+8}{-4}=\frac{-24}{-4}=6\)