rút gọn biểu thức : A=(3x3+3x+1).(3x3-3x+1)-(3x3+1)2
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\(A=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left[\left(3x^3+1\right)+3x\right]\left[\left(3x^3+1\right)-3x\right]-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-\left(3x\right)^2-\left(3x^3+1\right)^2\)
\(=-\left(3x\right)^2\)
\(=-9x^2\)
a, Với x khác 1
\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)
b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)
Vậy với x khác 1 thì bth A luôn nhận gtri âm
a: F(x)=3x^3-2x^2+5x-7
G(x)=3x^3-2x^2+5x+7x^2+3=3x^3+5x^2+5x+3
Bậc của F(x),G(x) đều là 3
b: N(x)=G(x)-F(x)
\(=3x^3+5x^2+5x+3-3x^3+2x^2-5x+7=7x^2+10\)
M(x)=2F(x)+G(x)
\(=6x^3-4x^2+10x-14+3x^3+5x^2+5x+3\)
\(=9x^3+x^2+15x-11\)
c: x^2-3x=0
=>x=0 hoặc x=3
\(M\left(0\right)=9\cdot0^3+0^2+15\cdot0-11=-11\)
\(M\left(3\right)=9\cdot3^3+3^2+15\cdot3-11=286\)
d: N(x)=7x^2+10>=10
Dấu = xảy ra khi x=0
Đề lỗi quá. Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
ĐK: \(3x\ne\pm y;x\ne0\)
A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)
= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)
Thay x = 1; y=2, ta có:
A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)
\(a,=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\\ b,=\dfrac{\left(x+y\right)^2-16}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
a, \(\left(a+5\right)^2+2\left(a+5\right)\left(\dfrac{1}{2}-a\right)+\left(\dfrac{1}{2}-a\right)^2=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\)
b,\(\dfrac{x^2-16+2xy+y^2}{3x^2-12x+3xy}=\dfrac{\left(x^2+2xy+y^2\right)-4^2}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
\(A=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left[\left(3x^3+1\right)+3x\right]\left[\left(3x^3+1\right)-3x\right]-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-\left(3x\right)^2-\left(3x^3+1\right)^2\)
\(=-\left(3x\right)^2\)
\(=-9x^2\)