ĐKXĐ: \(\left\{{}\begin{matrix}0\le x\le\frac{1}{2}\\0\le y\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow xy\le\frac{1}{4}\)

Từ pt đầu: \(\Leftrightarrow\frac{4}{1+2xy}=\left(\frac{1}{\sqrt{1+2x^2}}+\frac{1}{\sqrt{1+2y^2}}\right)^2\le2\left(\frac{1}{1+2x^2}+\frac{1}{1+2y^2}\right)\)

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

\(\Leftrightarrow\frac{1}{1+2x^2}+\frac{1}{1+2y^2}-\frac{2}{1+2xy}\ge0\)

\(\Leftrightarrow\frac{2\left(2xy-1\right)\left(x-y\right)^2}{\left(1+2x^2\right)\left(1+2y^2\right)\left(1+2xy\right)}\ge0\) (2)

Do \(xy\le\frac{1}{4}< \frac{1}{2}\Rightarrow2xy-1< 0\)

\(\Rightarrow\left(2\right)\) xảy ra khi và chỉ khi \(x-y=0\Leftrightarrow x=y\)

Thế vào pt dưới:

\(2\sqrt{x\left(1-2x\right)}=\frac{2}{9}\Leftrightarrow x\left(1-2x\right)=\frac{1}{81}\Leftrightarrow...\)

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$

\(y\left(x+1\right)^2=-x^2+2018x-1\)

\(\Leftrightarrow y=\dfrac{-x^2+2018x-1}{\left(x+1\right)^2}=-1+\dfrac{2020x}{\left(x+1\right)^2}\)

\(\Rightarrow\dfrac{2020x}{\left(x+1\right)^2}\in Z\)

Mà x và \(x\left(x+2x\right)+1\) nguyên tố cùng nhau

\(\Rightarrow2020⋮\left(x+1\right)^2\)

Ta có 2020 chia hết cho đúng 2 số chính phương là 1 và 4

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=1\\\left(x+1\right)^2=4\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow y\)

b.

Từ pt đầu:

\(x^2+xy-2y^2+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y-2\end{matrix}\right.\)

Thế xuống dưới ...

Điều kiện: \(x,y\le\frac{1}{2}\Rightarrow2xy\le\frac{1}{2}\)

Ta có: 

\(\left(\frac{1}{\sqrt{1+2x^2}}+\frac{1}{\sqrt{1+2y^2}}\right)^2\le2\left(\frac{1}{1+2x^2}+\frac{1}{1+2y^2}\right)\)

\(\le\frac{4}{1+2xy}\)

\(\Rightarrow x=y\)

Làm nốt

[1=71]

(1=71)OK BẠN

1/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=4\\\frac{1}{\left(x+1\right)^2-1}+\frac{1}{\left(y+1\right)^2-1}=\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4\\\frac{1}{\left(x+1\right)^2-1}+\frac{1}{\left(y+1\right)^2-1}=\frac{2}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=a\\y+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=4\\\frac{1}{a^2-1}+\frac{1}{b^2-1}=\frac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\frac{1}{a^2-1}+\frac{1}{\frac{16}{a^2}-1}=\frac{2}{3}\)

\(\Rightarrow a^4-8a^2+16=0\Rightarrow a^2=4\Rightarrow a=\pm2\Rightarrow x=...\)

b/ ĐKXĐ: ...

\(\Rightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)

\(\Rightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)

\(\Rightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\sqrt{2-\frac{1}{y}}+xy\sqrt{2-\frac{1}{x}}}=0\)

\(\Rightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\sqrt{2-\frac{1}{y}}+xy\sqrt{2-\frac{1}{x}}}=0\right)\)

\(\Rightarrow\sqrt{y}=\sqrt{x}\Rightarrow y=x\) (ngoặc phía sau luôn dương)

Thay vào pt đầu:

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

Mặt khác áp dụng BĐT \(a+b\le\sqrt{2\left(a^2+b^2\right)}\)

\(\Rightarrow\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}\le\sqrt{2\left(\frac{1}{x}+2-\frac{1}{x}\right)}=2\)

Dấu "=" xảy ra khi và chỉ khi:

\(\frac{1}{\sqrt{x}}=\sqrt{2-\frac{1}{x}}\Rightarrow\frac{1}{x}=2-\frac{1}{x}\Rightarrow x=1\Rightarrow y=1\)

1/ \(\Leftrightarrow\left\{{}\begin{matrix}2x^2-4xy+2x-4y+6=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2-2xy+4x-4y+4=0\)

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

\(\Leftrightarrow\left(x-y+2\right)^2=0\)

\(\Rightarrow y=x+2\)

Thay vào 1 trong 2 pt ban đầu là xong

2/ \(x^2-\left(y+2\right)x-6y^2+11y-3=0\)

\(\Delta=\left(y+2\right)^2-4\left(-6y^2+11y-3\right)\)

\(=25y^2-40y+16=\left(5y-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{y+2+5y-4}{2}\\x=\frac{y+2-5y+4}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3y-1\\x=-2y+3\end{matrix}\right.\)

Thay vào pt 2 là được

c/ \(S=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{100}}\)

\(S< 1+\frac{2}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

\(S< 1+2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(S< 1+2\left(\sqrt{100}-1\right)=19\)

\(S>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{101}-\sqrt{100}}\)

\(S>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)

\(S>2\left(\sqrt{101}-1\right)>2\left(\sqrt{100}-1\right)=18\)

\(\Rightarrow18< S< 19\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải số tự nhiên

a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)