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Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)
\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)
\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)
\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)
Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
1/ \(x-1=\sqrt[3]{2}\Rightarrow\left(x-1\right)^3=2\Rightarrow x^3-3x^2+3x-3=0\)
\(B=x^2\left(x^3-3x^2+3x-3\right)+x\left(x^3-3x^3+3x-3\right)+x^3-3x^2+3x-3+1945\)
\(B=1945\)
b/ Tương tự:
\(x-1=\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x^3-3x^2+3x-1=6+3\sqrt[3]{8}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Rightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)
\(\Rightarrow x^3-3x^2-3x-1=0\)
\(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)
\(P=2016\)
1/\(A=\dfrac{x^2-2x+2014}{x^2}\)
\(\Leftrightarrow A=\dfrac{2014x^2-2.x.2014+2014^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013x^2+x^2-2.x.2014+2014^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)
\(\Leftrightarrow A=\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\)
Có: \(\left(x-2014\right)^2\ge0\forall x\)
\(2014x^2>0\forall xvìx\ne0\)
\(\Rightarrow\dfrac{\left(x-2014\right)^2}{2014x^2}\ge0\)
\(\Rightarrow\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\ge\dfrac{2013}{2014}\)
\(\Rightarrow A\ge\dfrac{2013}{2014}\)
dấu "=" xảy ra khi và chỉ khi x - 2014 =0 <=> x = 2014
Vậy \(min_A=\dfrac{2013}{2014}\Leftrightarrow x=2014\)
2) Ta có:
\(x=\sqrt{a+\sqrt{a^2-1}}+\sqrt{a-\sqrt{a^2-1}}\)
\(\Leftrightarrow x^2=a-\sqrt{a^2-1}+2\sqrt{a-\sqrt{a^2-1}}.\sqrt{a+\sqrt{a^2-1}}+a+\sqrt{a^2-1}\)
\(\Leftrightarrow x^2=2a+2.\sqrt{\left(a-\sqrt{a^2-1}\right)\left(a+\sqrt{a^2-1}\right)}\)
\(\Leftrightarrow x^2=2a+2\sqrt{a^2-\left(a^2-1\right)}\)
\(\Leftrightarrow x^2=2a+2=2\left(a+1\right)\)
\(\Leftrightarrow-x^3=-2\left(a+1\right)x\)
Đặt \(A=x^3-2x^2-2\left(a+1\right)x+4x+2021\)
\(\Leftrightarrow A=x^3-2\left(2a+2\right)-x^3+4a+2021\)
\(\Leftrightarrow A=-4a-4+4a+2021\)
\(\Leftrightarrow A=2017\)
Đề lỗi rồi. Bạn xem lại đề.