\(y\times\dfrac{6}{11}+y\times\dfrac{5}{11}=2025\)

\(y\times\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=2025\)

\(y\times1=2025\)

\(y=2025\)

\(y.\dfrac{6}{11}+y.\dfrac{5}{11}=2025\)

\(y.\left(\dfrac{6}{11}+\dfrac{5}{11}\right)=2025\)

\(y.\left(\dfrac{11}{11}\right)=2025\)

\(y=2025\)

Dấu . là dấu nhân

\(Y\times6+11\times\dfrac{5}{11}=2025\\ Y\times6+5=2025\\ Y\times6=2025-5\\ Y\times6=2020\\ Y=\dfrac{2020}{6}\\ Y=\dfrac{1010}{3}\)

\(y\times6+11\times\dfrac{5}{11}=2025\\ y\times6+5=2025\\y\times6=2025-5\\ y\times6=2020\\ y=2020:6\\ y=\dfrac{1010}{3}\)

\(\begin{cases}x-\left|y-5\right|=8\\ \left|x+11\right|+3\left|y-5\right|=21\end{cases}\)

=>\(\begin{cases}3x-3\left|y-5\right|=24\\ \left|x+11\right|+3\left|y-5\right|=21\end{cases}\)

=>3x-3|y-5|+|x+11|+3|y-5|=24+21

=>3x+|x+11|=45(1)

TH1: x>=-11

(1) sẽ trở thành: 3x+x+11=45

=>4x=34

=>x=8,5(nhận)

x-|y-5|=8

=>|y-5|=x-8=8,5-8=0,5

=>\(\left[\begin{array}{l}y-5=0,5\\ y-5=-0,5\end{array}\right.=>\left[\begin{array}{l}y=5,5\\ y=4,5\end{array}\right.\)

TH2: x<11

(1) sẽ trở thành: 3x-x-11=45

=>2x=45+11=56

=>x=28(loại)

=>Loại

Lời giải:
ĐKXĐ: $x\geq 0; y\geq 1; z\geq 2$

PT \(\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-4\sqrt{y-1}+4]+[(z-2)-6\sqrt{z-2}+9]=0\)

\(\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-2)^2+(\sqrt{z-2}-3)^2=0\)

Vì \((\sqrt{x}-1)^2, (\sqrt{y-1}-2)^2, (\sqrt{z-2}-3)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:
\((\sqrt{x}-1)^2=(\sqrt{y-1}-2)^2=(\sqrt{z-2}-3)^2=0\)

$\Leftrightarrow x=1; y=5; z=11$

Ta có: \(\begin{cases} \sqrt{11}x + \sqrt{5}y = 16 & (1) \\ 3x - 2y = -2\sqrt{5} + 3\sqrt{11} & (2) \end{cases}\)

=>\(\begin{cases}2\sqrt{11}x+2\sqrt{5}y=32\quad\\ 3\sqrt{5}x-2\sqrt{5}y=\sqrt{5}(-2\sqrt{5}+3\sqrt{11})=-10+3\sqrt{55}\quad\end{cases}\)

=>\(\begin{cases}(2\sqrt{11}+3\sqrt{5})x=32-10+3\sqrt{55}\\ 3x-2y=-2\sqrt5+3\sqrt{11}\end{cases}\)

=>\(\begin{cases}(2\sqrt{11}+3\sqrt{5})x=22+3\sqrt{55}\\ 3x-2y=-2\sqrt5+3\sqrt{11}\end{cases}\Rightarrow\begin{cases}x=\sqrt{11}\\ 2y=3\sqrt{11}+2\sqrt5-3\sqrt{11}=2\sqrt5\end{cases}\)

=>\(\begin{cases}x=\sqrt{11}\\ y=\sqrt5\end{cases}\)

a: \(\sqrt{x^2+6x+9}=\sqrt{11+6\sqrt{2}}\)

=>\(\sqrt{\left(x+3\right)^2}=\sqrt{\left(3+\sqrt{2}\right)^2}\)

=>\(\left|x+3\right|=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)

=>\(\left[{}\begin{matrix}x+3=3+\sqrt{2}\\x+3=-3-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-6-\sqrt{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=4\\x+2y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-2y=8\\x+2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y+x+2y=8-3\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=5\\y=2x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\cdot1-4=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-y=-9\\x+y=11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=2\\x+y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=11-x=10\end{matrix}\right.\)

Giải hệ phương trình:
\(\left\{{}\begin{matrix}x-y=-9\\x+y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-9+y\\-9+y+y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-9+y\\2y-9=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-9+y\\2y=20\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-9+10\\y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=10\end{matrix}\right.\)

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

Vậy hệ phương trình đã cho có nghiệm (x; y) = (56; -9).