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a) \(\sqrt{9-12x+4x^2}=4+x\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
th1: \(3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow\Leftrightarrow x\le\dfrac{3}{2}\)
\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow3-2x=4+x\Leftrightarrow3x=-1\Leftrightarrow x=\dfrac{-1}{3}\left(tmđk\right)\)
th2: \(3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}\)
\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow2x-3=4+x\Leftrightarrow x=7\left(tmđk\right)\)
vậy \(x=\dfrac{-1}{3};x=7\)
b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)
\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)
\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)
th1: \(2-x\ge0\Leftrightarrow x\le2\)
\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow2-x=x^2-x-5\)
\(\Leftrightarrow x^2=7\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{7}\left(loại\right)\\x=-\sqrt{7}\left(tmđk\right)\end{matrix}\right.\)
th2: \(2-x< 0\Leftrightarrow x>2\)
\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow x-2=x^2-x-5\)
\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(tmđk\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
vậy \(x=-\sqrt{7};x=3\)
a) \(\sqrt{9-12x+4x^2}=4+x\)
\(\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x=4+x\\3-2x=-4-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=7\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{3};x_2=7\).
b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)
\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)
\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=x^2-x-5\\2-x=-x^2+x+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=7\\x^2=2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\left(l\right)\\x=-\sqrt{7}\\x=3\\x=-1\left(l\right)\end{matrix}\right.\)
Vậy \(x_1=-\sqrt{7};x_2=3\).
a, \(\Leftrightarrow\sqrt{\left(3-2x\right)^2=4+x}\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=4+x\\3-2x=-4-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\\\left(x-3\right)\left(x-1\right)=0\end{cases}}\)
:V
Câu đầu cho x > 0 thì dễ hơn ......
Sử dụng BĐT AM - GM ta dễ có:\(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\ge2\sqrt{\left(\sqrt{x}+2\right)\cdot\frac{9}{\sqrt{x}+2}}-2=4\)
Đẳng thức xảy ra tại x=1
\(E=\frac{x+1}{\sqrt{x}}\ge\frac{2\sqrt{x}}{\sqrt{x}}=2\) Đẳng thức xảy ra tại x=1
Làm 2 cái thôi còn lại tương tự bạn nhé :)
+ Ta có: \(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}\)
\(D=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\)
Áp dụng bất đẳng thức Cô-si cho phương trình \(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\) ta có:
\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge\sqrt{\left(\sqrt{x}+2\right).\left(\frac{9}{\sqrt{x}+2}\right)}=\sqrt{9}=3\)
\(\Rightarrow\)\(D\ge3-2=1\)
Dấu bằng xảy ra khi và chỉ khi: \(\sqrt{x+2}=\frac{9}{\sqrt{x}+2}\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)^2=9\)
\(\Leftrightarrow\sqrt{x}+2=\pm3\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=-3\\\sqrt{x}+2=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-5\left(L\right)\\\sqrt{x}=1\end{cases}}\)
\(\Leftrightarrow x=\pm1\)
Vậy \(S=\left\{\pm1\right\}\)
a)\(\sqrt{x}+1>\sqrt{x+1}\) (x>0)
Có:\(\left(\sqrt{x}+1\right)^2=x+2\sqrt{x}+1\left(1\right)\) (x>0)
\(\sqrt{\left(x+1\right)^2}=x+1\) (2) (x>0)
từ (1) và (2) =>(đpcm)
b)\(\sqrt{x^2+1}>x\)
Có:\(\sqrt{\left(x^2+1\right)^2}=x^2+1\left(1\right)\)
x2=x2 (2)
Từ (1) và (2) =>(đpcm)
c)\(\frac{1}{2}+a+b\ge\sqrt{a}+\sqrt{b}\left(a,b\ge0\right)\)
Vì a,b >or= 0
=>\(a+b\ge\sqrt{a}+\sqrt{b}\)
\(\Rightarrow\frac{1}{2}+a+b\ge\sqrt{a}+\sqrt{b}\) (đáng lẽ 1/2+a+b> mới phải)
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1\)(*)
Vì \(x\ge2\Rightarrow x-1\ge1\Rightarrow\sqrt{x-1}\ge1\Rightarrow\sqrt{x-1}-1\ge0\)
Khi đó (*)\(=\sqrt{x-1}-1+\sqrt{x-1}+1=2\sqrt{x-1}\)(đpcm)