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\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)
\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)
\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)
\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)
\(=\left(-1\right)^{2018}+2018=2019\)
\(C=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2=13-ab\)
Có : \(a^2+2ab+b^2=\left(a+b\right)^2\)
\(\Leftrightarrow13+2ab=1^2\)
\(\Leftrightarrow2ab=-12\)
\(\Leftrightarrow ab=-6\)
\(\Leftrightarrow C=16-\left(-6\right)=13+6=19\)
\(a+b=1\Rightarrow\left(a+b\right)^2=1\)\(\Rightarrow a^2+b^2+2ab=1\)
Mà \(a^2+b^2=13\Rightarrow ab=-6\)
\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=1.\left[13-\left(-6\right)\right]=19\)
\(D=\left(x-1\right)^2-\left(2x+1\right)^2+2018\)
\(=x^2-2x+1-4x^2-4x-1+2018\)
\(=-3x^2-6x+2018=-3\left(x^2+2x+1\right)+2021\)
\(=-3\left(x+1\right)^2+2021\)
Vì \(-3\left(x+1\right)^2\le0\Rightarrow D\le2021\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-1\)
Vậy \(Max_D=2021\Leftrightarrow x=-1\)
\(D=\left(\sqrt{2}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{2}-1\right)\\ D=\left(\sqrt{10}-\sqrt{5}+\sqrt{2}-1\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{2}-1\right)\\ D=\left(\sqrt{10}-1\right)^2-\left(\sqrt{5}-\sqrt{2}\right)^2\\ D=10-2\sqrt{10}+1-5+2\sqrt{10}-2\\ D=4\)
a, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
\(A=\sqrt{2}\left(\sqrt{2}-2\right)+\left(\sqrt{2}+1\right)^2\)
\(A=2-2\sqrt{2}+2+2\sqrt{2}+1\)
\(A=5\)
\(A=\sqrt{2}\left(\sqrt{2}-2\right)+\left(\sqrt{2}+1\right)^2\\ A=2-2\sqrt{2}+2+2\sqrt{2}+1\\ A=5\)