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Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
a,ĐKXĐ:\(\left\{{}\begin{matrix}x-4\ne0\\x+4\ne0\\x^2-16\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-4\\x\ne\pm4\end{matrix}\right.\Leftrightarrow x\ne\pm4\)
b,\(\dfrac{4}{x-4}+\dfrac{3}{x+4}.\dfrac{6x}{x^2-16}=\dfrac{4}{x-4}+\dfrac{18x}{\left(x-4\right)\left(x+4\right)^2}=\dfrac{4\left(x+4\right)^2+18x}{\left(x-4\right)\left(x+4\right)^2}=\dfrac{4\left(x^2+8x+16\right)+18x}{\left(x-4\right)\left(x+4\right)^2}=\dfrac{4x^2+32x+64+18x}{\left(x-4\right)\left(x+4\right)^2}=\dfrac{4x^2+50x+64}{\left(x-4\right)\left(x+4\right)^2}\)
1/2 x 2 (6x – 3) – x( x 2 + 1/2) + 1/2.(x + 4)
= (3 x 3 – 3/2. x 2 ) – ( x 3 + 1/2.x) + (1/2.x + 2)
= 3 x 3 - 3/2 x 2 – x 3 - 1/2 x + 1/2 x + 2
= ( 3 x 3 – x 3 ) - 3/2. x 2 – (1/2 x - 1/2 x) + 2
= 2 x 3 - 3/2 x 2 + 2
a: \(A=\dfrac{x^4+x^2+11x^2+11}{x^4+x^2+5x^2+5}=\dfrac{\left(x^2+11\right)\left(x^2+1\right)}{\left(x^2+5\right)\left(x^2+1\right)}=\dfrac{x^2+11}{x^2+5}\)
b: \(A=\dfrac{x^2+5+6}{x^2+5}=1+\dfrac{6}{x^2+5}< =1+\dfrac{6}{5}=\dfrac{11}{5}\)
Dấu = xảy ra khi x=0
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
\(a,\dfrac{x^2+6x+9}{x+3}\\ đk:x\ne-3\\ =\dfrac{\left(x+3\right)^2}{x+3}=x+3\)
b, Thay \(x=-2\left(t/mđk\right)\) vào
\(-2+3=1\)
Vậy tại \(x=-2\) thì biểu thức = 1
\(A=\dfrac{x^2+6x+9}{x+3}\)
\(A=\dfrac{x^2+2.x.3+3^2}{x+3}\)
\(A=\dfrac{\left(x+3\right)^2}{x+3}\)
\(A=x+3\)
b) Thay x = -2 vào A ta được A = -2 + 3 = 1
Vậy khi x = -2 thì A = 1
Khi x=2 thì A=4+2+12=18