Điều kiện: \(\orbr{\begin{cases}x\ge1\\x\le-4-\sqrt{7}\end{cases}}\)

Đặt \(\hept{\begin{cases}\sqrt{2x^2+16x+18}=a\\\sqrt{x^2-1}=b\end{cases}}\left(a,b\ge0\right)\)

Phương trình trở thành:

\(a+b=\sqrt{a^2+2b^2}\)

\(\Leftrightarrow a^2+2ab+b^2=a^2+2b^2\)

\(\Leftrightarrow b^2-2ab=0\Leftrightarrow\orbr{\begin{cases}b=0\\b=2a\end{cases}}\)

+) Nếu \(b=0\Rightarrow\sqrt{x^2-1}=0\Leftrightarrow x=1\)(do ĐK)

+) Nếu \(b=2a\Rightarrow\sqrt{x^2-1}=2\sqrt{2x^2+16x+18}\)

\(\Leftrightarrow x^2-1=4\left(2x^2+16x+18\right)\)

\(\Leftrightarrow7x^2+64x+73=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-32+3\sqrt{57}}{7}\left(L\right)\\x=\frac{-32-3\sqrt{57}}{7}\left(C\right)\end{cases}}\)

Vậy \(S=\left\{1;\frac{-32-3\sqrt{57}}{7}\right\}\)

Ta có: \(Q=\frac{x+y}{\sqrt{3x^2-xy+y^2}}=\frac{1+\frac{y}{x}}{\sqrt{3-\frac{y}{x}+\left(\frac{y}{x}\right)^2}}=\frac{1+t}{\sqrt{3-t+t^2}}\left(t=\frac{y}{x}\right)\)

Từ xy+1\(\le x\Rightarrow\frac{y}{x}+\frac{1}{x^2}\le\frac{1}{x}\Rightarrow\frac{y}{x}\le\frac{1}{x}-\frac{1}{x^2}=\frac{1}{4}-\left(\frac{1}{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\Rightarrow0\le t\le\frac{1}{4}\)

Vì \(t\le\frac{1}{4}\Rightarrow1+t\le\frac{5}{4};3-t+t^2=\left(\frac{1}{2}-t\right)^2+\frac{11}{4}\ge\frac{45}{16}\)

\(\Rightarrow Q\le\frac{\frac{5}{4}}{\sqrt{\frac{45}{16}}}=\frac{\sqrt{5}}{3}\)

Dấu đẳng thức xảy ra <=>  \(\hept{\begin{cases}t=\frac{1}{4}\\x=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy GTLN của Q là \(\frac{\sqrt{5}}{3}\)

Hoành độ giao điểm thỏa mãn phương trình

\(x^2+4=5x-2\Leftrightarrow x^2-5x+6=0\)

\(\Delta=25-24=1>0\)

\(x_1=\frac{5-1}{2}=2;x_2=\frac{5+1}{2}=3\)

Với \(x=2\Rightarrow y=5.2-2=8\)

Với \(x=3\Rightarrow y=5.3-2=13\)

Vậy giao điểm của (p) cà (d) có tọa độ là A ( 2 ; 8 ) ; B ( 3 ; 13 ) 

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)

=> đpcm . Dấu "=" xảy ra <=> x = y = 1/2

Đặt \(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(=x^2+\frac{1}{x^2}+2+y^2+\frac{1}{y^2}+2\)

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

Áp dụng BĐT Co-si , có :

\(x^2+y^2\ge2xy\)

\(\Rightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Có \(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\ge\frac{8}{\left(x+y\right)^2}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}\ge8\)

\(\Rightarrow A\ge\frac{1}{2}+8+4\)

\(\Rightarrow A\ge\frac{25}{2}\)

ĐK: \(-2\le x\le2\).

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

\(t^2=4+2\sqrt{4-x^2}\ge4\)

Dấu \(=\)khi \(x=\pm2\).

\(t^2=4+2\sqrt{4-x^2}\le4+2\sqrt{4}=8\)

Dấu \(=\)khi \(x=0\).

Suy ra \(4\le t^2\le8\Leftrightarrow2\le t\le2\sqrt{2}\).

\(P=t-\frac{t^2-4}{2}\Leftrightarrow2P=-t^2+2t+4=-\left(t-1\right)^2+5\)

Vì \(2\le t\le2\sqrt{2}\)nên \(\hept{\begin{cases}2P\ge-\left(2\sqrt{2}-1\right)^2+5=-4+4\sqrt{2}\\2P\le-\left(2-1\right)^2+5=4\end{cases}}\)

\(\Rightarrow P\le2\)dấu \(=\)khi \(t=2\)\(\Rightarrow x=\pm2\).

\(P\ge2\sqrt{2}-2\)dấu \(=\)khi \(t=2\sqrt{2}\)\(\Rightarrow x=0\).

\(B=\frac{23-4\sqrt{5}-3}{\sqrt{23-4\sqrt{5}-2}-1}\)

\(=\frac{20-4\sqrt{5}}{\sqrt{21-4\sqrt{5}}-1}=\frac{20-4\sqrt{5}}{\sqrt{\left(2\sqrt{5}-1\right)^2}-1}\)

\(=\frac{20-4\sqrt{5}}{2\sqrt{5}-1-1}=\frac{20-4\sqrt{5}}{2\sqrt{5}-2}=\frac{10-2\sqrt{5}}{\sqrt{5}-1}\)

đk: \(x,y\ge-6\Rightarrow x+y\ge0\)

Theo bài ra, ta có: 

\(\left(x+y\right)^2=\left(\sqrt{x+6}+\sqrt{y+6}\right)^2\)

\(=x+y+12+2\sqrt{\left(x+6\right)\left(y+6\right)}\le x+y+12+x+6+y+6\)

Hay \(\left(x+y\right)^2\le2\left(x+y\right)+24\)

\(\Leftrightarrow\left(x+y+4\right)\left(x+y-6\right)\le0\)

\(\Leftrightarrow x+y-6\le0\Leftrightarrow x+y\le6\)

Dấu '=' xảy ra<=> x=y=3

=> GTNN của P là 6 <=> x=y=3

Đặt \(a=\sqrt{x+6};b=\sqrt{y+6}\Rightarrow a;b\ge0,a+b=a^2+b^2-12\)

và \(P=a^2+b^2-12=a+b\)

Ta có: \(a+b=\left(a+b\right)^2-2ab-12\Rightarrow a+b\le\left(a+b\right)^2-12\left(a;b\ge0\right)\)

Hay \(\left(a+b\right)^2-\left(a+b\right)-12\ge0\)

\(\Leftrightarrow\left(a+b+3\right)\left(a+b-4\right)\ge0\)

\(\Leftrightarrow a+b\ge4\Rightarrow P\ge4\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=4\\b=0\end{cases}}\) hoặc \(\hept{\begin{cases}a=0\\b=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=10\\y=-6\end{cases}}\) hoặc \(\hept{\begin{cases}x=-6\\y=10\end{cases}}\)

ĐK: \(0\le x\le1\).

Ta có: 

với \(0\le x\le1\)thì \(0\le1-x\le1\Leftrightarrow\sqrt{1-x}\left(1-\sqrt{1-x}\right)\ge0\Leftrightarrow\sqrt{1-x}\ge1-x\)

do đó \(x+\sqrt{1-x}\ge x+1-x=1\Rightarrow\sqrt{x+\sqrt{1-x}}\ge1\).

Do đó để phương trình đã cho có nghiệm thì: 

\(\hept{\begin{cases}\sqrt{x}=0\\x+\sqrt{1-x}=1\end{cases}}\Leftrightarrow x=0\left(tm\right)\).

\(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)

\(\Leftrightarrow x^3-5x^2+11x-7=6\sqrt[3]{x^2-x+1}-3x-3\)

\(\Leftrightarrow x^3-5x^2+11x-7=3\frac{8x^2-8x+8-\left(x^3+3x^2+3x+1\right)}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\)

\(\Leftrightarrow\left(x^3-5x^2+11x-7\right)\left(1+\frac{3}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\right)=0\)

\(\Leftrightarrow x^3-5x^2+11x-7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+7\right)=0\)

\(\Leftrightarrow x=1\).

cảm ơn anh đã giúp ạ

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

\(\Leftrightarrow x-2=16-8x+x^2\)

\(\Leftrightarrow x^2-9x+18=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

\(x-4+\sqrt{x-2}=0\)

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

bình phương 2 vế : \(\left|x-2\right|=\left(4-x\right)^2=x^2-8x+16\)

ĐK : \(\left(4-x\right)^2\ge0\Leftrightarrow x\le4\)

TH1 : \(x-2=x^2-8x+16\Leftrightarrow x^2-9x+18=0\)

\(\Delta=81-4.18=9>0\)

\(x_1=\frac{9-3}{2}=3\left(tm\right);x_2=\frac{9+3}{2}=6\left(ktm\right)\)

TH2 : \(-x+2=x^2-8x+16\Leftrightarrow x^2-7x+14=0\)

\(\Delta=49-4.14< 0\)phương trình vô nghiệm 

Vậy tập nghiệm của phương trình là S = { 3 }