Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
1. \(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|=1\)
+ Ta có : \(\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)
Dấu "=" \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x}\le3\Leftrightarrow4\le x\le9\)
2. + \(ĐK:4-2x-x^2\ge0\)
+ VT = \(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\) \(\ge\sqrt{4}+\sqrt{9}=5\) (1)
Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
+ VP \(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\forall x\) (2)
Dấu "=" \(\Leftrightarrow x=-1\)
+ Từ (1) và (2) suy ra : pt \(\Leftrightarrow VT=VP=5\Leftrightarrow x=-1\) (TM)
3. + TH1: \(x< 0\) ta có :
\(VT< \sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
+ TH2 : x = 0 ta có :
\(VT=\sqrt[3]{1}+\sqrt[3]{0}=1\) ( TM )
+ TH3 : x > 0 ta có :
\(VT>\sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
Vậy x = 0 là nghiệm duy nhất của pt
4. \(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+3\right)-24=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x-8\right)-24=0\)
\(\Leftrightarrow t\left(t-5\right)-24=0\) ( với \(t=x^2+2x-3\) )
\(\Leftrightarrow t^2-5t-24=0\Leftrightarrow\left(t+3\right)\left(t-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-3\\x^2+2x-3=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\sqrt{3}-1\\x=-2\sqrt{3}-1\end{matrix}\right.\) ( TM )
ĐKXĐ : x\(\ge0\)
ADBĐT BCS ta được
\(\left(\frac{x^2}{3}+4\right)\left(3+1\right)\ge\left(x+2\right)^2\)
\(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge2x+4\)(do x\(\ge0\)) (1)
Do x\(\ge0\)nên ADBĐT Cauchy ta được:
\(\sqrt{6x}\le\frac{x+6}{2}\)\(\Rightarrow1+\frac{3x}{2}+\sqrt{6x}\le1+\frac{3x}{2}+\frac{x+6}{2}=1+\frac{4x+6}{2}=2x+4\)(2)
Từ (1) và (2) \(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge1+\frac{3x}{2}+\sqrt{6x}\)
Dấu = xảy ra \(\Leftrightarrow x=6\)(thỏa mãn ĐKXĐ)
3) ĐKXĐ \(-1\le x\le1\)
Khi đó phương trình đã cho \(\Leftrightarrow4\left(\sqrt{1+x}+\sqrt{1-x}\right)=8-x^2\)
\(\Leftrightarrow\hept{\begin{cases}16\left(2+2\sqrt{1-x^2}\right)=\left(7+1-x^2\right)\left(2\right)\\8-x^2\ge0\end{cases}}\)
Đặt \(\sqrt{1-x^2}=a\ge0\)
Khi đó phương trình (2) trở thành:
\(\hept{\begin{cases}16\left(2+2a\right)=\left(7+a^2\right)\\x^2\le8\end{cases}}\)
\(\Leftrightarrow a^4+14a^2+49=32+32a\)
\(\Leftrightarrow a^4+14a^2-32a+17=0\)
\(\Leftrightarrow a^4-2a^2+1+16a^2-32a+16=0\)
\(\Leftrightarrow\left(a^2-1\right)^2+16\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
hay \(\sqrt{1-x^2}=1\)
\(\Leftrightarrow x=0\)(thỏa mãn)
