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\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)
\(=\sqrt{\left(x^2-x-2\right)\left(x^2-x\right)+5}\)
Đặt \(t=x^2-x\) ta đc:
\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)
\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)
Vậy....
b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=\left|x-2\right|+\left|x+3\right|\)
Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)
Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)
\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
1) ĐK: \(x\ge-2012\)
Đặt \(\sqrt{x+2012}=t\left(t\ge0\right)\Rightarrow x=t^2-2012\)
Ta có hệ \(\hept{\begin{cases}x^2+t=2012\\-x+t^2=2012\end{cases}}\)
\(\Rightarrow x^2+t-t^2+x=0\Rightarrow\left(x+t\right)\left(x-t+1\right)=0\)
Với \(x+t=0\Leftrightarrow\sqrt{x+2012}=x\Rightarrow x^2-x-2012=0\Rightarrow x=\frac{\sqrt{8049}+1}{2}\)
Với \(x-t+1=0\Leftrightarrow\sqrt{x+2012}=x+1\Rightarrow x^2+x-2011=0\Rightarrow x=\frac{\sqrt{8045}-1}{2}\)
2) ĐK \(\orbr{\begin{cases}x< -\frac{1}{3}\\x>1\end{cases}}\)
Đặt \(\sqrt{\frac{3x+1}{x-1}}=t\), phương trình trở thành \(4t+\frac{1}{t}=4\Rightarrow\frac{4t^2-4t+1}{t}=0\Rightarrow t=\frac{1}{2}\)
Khi đó ta có \(\sqrt{\frac{3x+1}{x-1}}=\frac{1}{2}\Rightarrow\frac{3x+1}{x-1}=\frac{1}{4}\Rightarrow11x+5=0\)
\(\Rightarrow x=-\frac{5}{11}\left(tm\right)\)
c) TH1: \(x\le-1\), phương trình trở thành \(\left(x-3\right)\left(x+1\right)-4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Đặt \(\sqrt{\left(x-3\right)\left(x+1\right)}=t\left(t\ge0\right)\) thì \(t^2-4t+3=0\Rightarrow\orbr{\begin{cases}t=1\\t=3\end{cases}}\)
Với \(t=1\Rightarrow\left(x-3\right)\left(x+1\right)=1\Rightarrow x^2-2x-4=0\Rightarrow\orbr{\begin{cases}x=1+\sqrt{5}\left(l\right)\\x=1-\sqrt{5}\left(tm\right)\end{cases}}\)
Với \(t=3\Rightarrow\left(x-3\right)\left(x+1\right)=9\Rightarrow x^2-2x-12=0\Rightarrow\orbr{\begin{cases}x=1+\sqrt{13}\left(l\right)\\x=1-\sqrt{13}\left(tm\right)\end{cases}}\)
Với \(x>3\), phương trình trở thành \(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Đặt \(\sqrt{\left(x-3\right)\left(x+1\right)}=t\left(t\ge0\right)\) thì \(t^2+4t+3=0\Rightarrow\orbr{\begin{cases}t=-1\\t=-3\end{cases}\left(l\right)}\)
Vậy pt có 2 nghiệm \(x=1-\sqrt{5}\) hoặc \(x=1-\sqrt{13}\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
\(a.\sqrt{\left(1-\sqrt{5}\right)^2}+1=\left|1-\sqrt{5}\right|+1=\sqrt{5}-1+1=\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}-2=\sqrt{\left(\sqrt{2}+1\right)^2}-2=\sqrt{2}+1-2=\sqrt{2}-1\)
\(c.\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)=\sqrt{\left(b-\dfrac{1}{2}\right)^2}-2b+\dfrac{1}{2}=b-\dfrac{1}{2}-2b+\dfrac{1}{2}=-2b\)
\(d.\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}\)
\(e.\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-2\)
\(g.3x+\sqrt{x^2-2x+1}=3x+\sqrt{\left(x-1\right)^2}\)
* \(x\ge1\Rightarrow3x+\left|x-1\right|=3x+x-1=4x-1\)
* \(x< 1\Rightarrow3x+\left|x-1\right|=3x+1-x=2x+1\)
\(h.\sqrt{y+2\sqrt{y^2-2y+1}}=\sqrt{y+2\sqrt{\left(y-1\right)^2}}=\sqrt{y+2y-2}=\sqrt{3y-2}\left(y\ge1\right)\) hoặc: \(\sqrt{y+2-2y}=\sqrt{-y+2}\left(y< 1\right)\)
\(H=\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
\(H^2=17-2\sqrt{32}+17+2\sqrt{32}+2\sqrt{\left(17-2\sqrt{32}\right)\left(17+2\sqrt{32}\right)}=34+2\sqrt{161}\)
\(H=\sqrt{34+2\sqrt{161}}\)
\(k.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)
b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)
\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)
c: \(=\sqrt{xy}+x\cdot\dfrac{\sqrt{y}}{\sqrt{x}}-y\cdot\dfrac{\sqrt{x}}{\sqrt{y}}\)
\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
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