ĐKXĐ: \(x>2;y>1\)

Khi đó Pt \(\Leftrightarrow\)\(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)

theo BĐT Cô si ta có \(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}\ge2.\sqrt{\frac{36}{\sqrt{x-2}}.4\sqrt{x-2}=24}\)

                                  và \(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4}{\sqrt{y-1}}.\sqrt{y-1}}=4\)

Pt đã cho có VT>= 28 Dấu "=" xảy ra \(\Leftrightarrow\)

\(\frac{36}{\sqrt{x-2}}=4\sqrt{x-2}\Leftrightarrow x=11\)

và \(\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\Leftrightarrow y=5\)

Đối chiếu với ĐK thì x=11; y=5 là nghiệm của PT

Ê Thắng tưởng off dòi mờ...nhanh thế....

ĐK: \(x\ge-1\)

\(\frac{pt\Leftrightarrow\sqrt{x+1}\sqrt{x^2-x+1}}{\sqrt{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)

\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=\sqrt{x^2-x+1}+\sqrt{x+3}\)

\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}=1\text{ (do }\sqrt{x^2-x+1}>0\text{)}\)

\(\Leftrightarrow...\)

\(DK:x\ge0\)

\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)

\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)

\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)

\(\Leftrightarrow x=1\)

Vay nghiem cua PT la \(x=1\)

Từ 1 đến 9 có số lượt chữ số là:

( 9 - 1 ) : 1 + 1 x 1 = 9 ( chữ số )

Từ 10 đến 99 có số lượt chữ số là:

[( 99 - 10 ) : 1 + 1 ] x 2 = 180 ( chữ số )

Từ 1 đến 100 có số lượt chữ số là:

180 + 9 + 3 = 192 ( chữ số )

Có 11 lượt chữ số 7 : 7;17;27;37;47;57;67;77;87;97

umgr hộ nha

xinlooix mình trả lời nhầm

\(VT=1.\sqrt{x}+2.\sqrt{x+3}\le\frac{x+1}{2}+\frac{2^2+x+3}{2}=x+4=VP\)

\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)

\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)

\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)

Làm nốt

ĐK: \(x^4-4x^3+14x-11\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)

e/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x+3-\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x-1}=x-1\)

\(\Leftrightarrow x-1=x^2-2x+1\)

\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)

a/ ĐKXĐ: ...

\(\Leftrightarrow9x+3\sqrt{x^2-x-1}=7x+7\)

\(\Leftrightarrow3\sqrt{x^2-x-1}=7-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\9\left(x^2-x-1\right)=\left(7-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\5x^2+19x-58=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\frac{29}{5}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1\)

\(\Leftrightarrow\frac{1}{\sqrt{\left(x-1\right)^2}}=\frac{1}{x-1}\)

\(\Leftrightarrow\frac{1}{\left|x-1\right|}=\frac{1}{x-1}\)

\(\Rightarrow x-1>0\Rightarrow x>1\)