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Ta có: \(x^2=\left(\sqrt{a+\sqrt{a^2-1}}+\sqrt{a-\sqrt{a^2-1}}\right)^2\)
\(=a+\sqrt{a^2-1}+2\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}+a-\sqrt{a^2-1}\)
\(=2a+2\sqrt{a^2-a^2+1}=2a+2=2\left(a+1\right)\)
Suy ra: \(x^3=x^2\cdot x=2\left(a+1\right)x\)
\(4a=2\cdot2a=2\left(2a+2\right)-4=2x^2-4\)
Nên \(P=x^3-2x^2-2\left(a+1\right)x+4a+2021\)
\(=x^3-2x^2-x^3+2x^2-4+2021=2021-4=2017\)
Nếu với \(1>a>0\)thì biểu thức dưới căn không xác định bạn nhé! =====> đề sai rồi.
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
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 \(A=x^3-2x^2-2\left(a+1\right)x+4x+2021\)
\(\Leftrightarrow A=x^3-2\left(2a+2\right)-x^3+4a+2021\)
\(\Leftrightarrow A=-4a-4+4a+2021\)
\(\Leftrightarrow A=2017\)
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)