= 168876548,8

do ban

\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)

\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)

Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :

\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)

\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)

\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)

\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)

sorry mn cho e sửa lại đề ạ

tìm gtln của p ạ

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)

\(x^3-3x+a⋮x^2-2x+1\)

\(\Leftrightarrow x^3-2x^2+x+2x^2-4x+2+a-2⋮x^2-2x+1\)

\(\Leftrightarrow a=2\)

\(2^{x-3}-3.2^x=-92\)

\(\Rightarrow2^x:2^3-3.2^x=-92\)

\(\Rightarrow2^x.\frac{1}{2^3}-3.2^x=-92\)

\(\Rightarrow2^x.\left(\frac{1}{2^3}-3\right)=-92\)

\(\Rightarrow2^x.\frac{-23}{8}=-92\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Trước hết bạn nhớ quy tắc phá dấu trị tuyệt đối này nhé :

\(\left|a\right|=\begin{cases}\text{a nếu }a\ge0\\\text{-a nếu }a< 0\end{cases}\)

Với \(3,5\le x\le4,1\)

thì : a/ \(\begin{cases}x-3,5\ge0\\4,1-x\ge0\end{cases}\)

\(\Rightarrow A=\left(x-3,5\right)+\left(4,1-x\right)=............\)

b/ \(\begin{cases}3,5-x\le0\\4,1-x\ge0\end{cases}\)

\(\Rightarrow B=-\left(-x+3,5\right)+\left(4,1-x\right)=.......................\)

^ la gi , roi minh se TL

x thuộc {-3; 3; -2; 2; -1; 1}

y thuộc {-5; 5; -4; 4; -3; 3}

Tìm x , y hả em 

| y + 3 | + ( x -2y )² \(\le\)0

Giải :

Vì | y + 3 | + ( x -2y )² \(\le\)0

Dấu " = " xảy ra khi 

\(\hept{\begin{cases}\left|y+3\right|=0\\\left(x-2y\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-3\\x=2y\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-3\\x=-6\end{cases}}\)

Vậy x = - 3 ; y = - 6