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

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

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

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

\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|\)

=2.

dấu = khi và chỉ khi \(\left(\sqrt{x-1}+1\right).\left(1-\sqrt{x-1}\right)=0\)

=0 nha bn

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

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

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

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

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

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

Dấu "=" xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\Leftrightarrow0\le x\le2\)

Vậy \(A_{min}=2\) tại \(0\le x\le2\)

adasdasdasd á d dá đâsdas

ĐK: \(x\ge1\)

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

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

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

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

Đẳng thức xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\)

\(\Leftrightarrow1\le x\le2\)

a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

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

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

b) \(\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow1\le x\le2\)

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

bn đang nằm mơ hay sao jậy

ừ :v t đang mơ đấy..nên đừng phá để t mơ tiếp

a) \(\orbr{\orbr{\begin{cases}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{cases}}}\)             b)\(\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\)

c)\(\orbr{\begin{cases}\hept{\begin{cases}x\ge\sqrt{2}\\x\ne\sqrt{3}\end{cases}}\\\hept{\begin{cases}x\le-\sqrt{2}\\x\ne-\sqrt{3}\end{cases}}\end{cases}}\)

Tìm x để căn có nghĩa ak mn giúp e với ak

\(a,ĐK:\dfrac{3}{x+7}\ge0\Leftrightarrow x+7>0\left(3>0;x+7\ne0\right)\Leftrightarrow x>-7\\ b,ĐK:\dfrac{-2}{5-x}\ge0\Leftrightarrow5-x< 0\left(2-< 0;5-x\ne0\right)\Leftrightarrow x>5\\ c,ĐK:x^2-7x+10\ge0\Leftrightarrow\left(x-5\right)\left(x-2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)

\(d,ĐK:x^2-8x+10\ge0\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4-\sqrt{6}\ge0\\x-4+\sqrt{6}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4-\sqrt{6}\le0\\x-4+\sqrt{6}\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4+\sqrt{6}\\x\ge4-\sqrt{6}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)

\(e,ĐK:9x^2+1\ge0\Leftrightarrow x\in R\left(9x^2+1\ge1>0\right)\)

a) \(ĐK:x+7>0\Leftrightarrow x>-7\)

b) \(ĐK:5-x< 0\Leftrightarrow x>5\)

c) \(ĐK:x^2-7x+10\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x-5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)

d) \(ĐK:x^2-8x+10\ge0\)

\(\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)

e) Do \(9x^2+1\ge1>0\)

Nên biểu thức được xác định với mọi x

\(\sqrt{x-2+2\sqrt{x+1}}+\sqrt{x+10+6\sqrt{x+1}}=2\sqrt{x+2+2\sqrt{x+1}}\)

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

Đặt \(y=\sqrt{x+1}\left(y\ge0\right)\)PT đã cho trở thành

\(y+1+\left|y-3\right|=2\left|y-1\right|\)

Nếu \(0\le y\le1:y+1+3-y=2-2y\Leftrightarrow y=-1\)(loại)

Nếu \(1\le y\le3:y+1+3-y=2y-2\Leftrightarrow y=3\)

Nếu y>3: y+1-y-3=2y-2 (vô nghiệm)

Với y=3 <=> x+1=9 <=> x=8

Vậy pt có 1 nghiệm x=8