Em kiểm tra lại câu a, chỗ \(x^2-x+z\) chữ \(z\) đó có vấn đề, nó phải là 1 con số ví dụ số 2 (chắc em nhìn nhầm số 2 thành chữ z)

Vâng em cảm ơn ạ!

Ta có: \(\begin{cases}\left(x-2\right)^2-2y^3=6\\ 3\left(x-2\right)^2+5y^3=7\end{cases}\)

=>\(\begin{cases}3\left(x-2\right)^2-6y^3=18\\ 3\left(x-2\right)^2+5y^3=7\end{cases}\Rightarrow\begin{cases}3\left(x-2\right)^2-6y^3-3\left(x-2\right)^2-5y^3=18-7\\ \left(x-2\right)^2-2y^3=6\end{cases}\)

=>\(\begin{cases}-11y^3=11\\ \left(x-2\right)^2=2y^3+6\end{cases}\Rightarrow\begin{cases}y^3=-1\\ \left(x-2\right)^2=2\cdot\left(-1\right)^3+6=4\end{cases}\)

=>\(\begin{cases}y=-1\\ x-2\in\left\lbrace2;-2\right\rbrace\end{cases}\Rightarrow\begin{cases}y=-1\\ x\in\left\lbrace4;0\right\rbrace\end{cases}\)

Đặt x+y=a; x-2y=b

=>6/a-3/b=3 và 1/a+7/b=2

=>a=5/3 và b=5

=>x+y=5/3 và x-2y=5

=>x=25/9; y=-10/9

a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

=>x+1=1 và y-2=1/2

=>x=0 và y=5/2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)

=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6

=>x-2y=9 và 2x-y=12

=>x=5; y=-2

c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

b: \(\begin{cases}5\left(x+2y\right)=3x-1\\ 2x+4=3\left(x-5y\right)-12\end{cases}\Rightarrow\begin{cases}5x+10y-3x=-1\\ 2x+4-3x+15y+12=0\end{cases}\)

=>\(\begin{cases}2x+10y=-1\\ -x+15y=-16\end{cases}\Rightarrow\begin{cases}2x+10y=-1\\ -2x+30y=-32\end{cases}\)

=>\(\begin{cases}2x+10y-2x+30y=-1-32\\ -x+15y=-16\end{cases}\Rightarrow\begin{cases}40y=-33\\ -x=-15y-16\end{cases}\)

=>\(\begin{cases}y=-\frac{33}{40}\\ x=15y+16=15\cdot\frac{33}{40}+16=\frac38\cdot33+16=\frac{227}{8}\end{cases}\)

a: \(\begin{cases}x\left(\sqrt5+2\right)+y=3-\sqrt5\\ -x+2y=6-2\sqrt5\end{cases}\Rightarrow\begin{cases}x\left(2\sqrt5+4\right)+2y=6-2\sqrt5\\ -x+2y=6-2\sqrt5\end{cases}\)

=>\(\begin{cases}x\left(2\sqrt5+4\right)+2y+x-2y=6-2\sqrt5-6+2\sqrt5=0\\ -x+2y=6-2\sqrt5\end{cases}\)

=>\(\begin{cases}x\left(2\sqrt5+5\right)=0\\ 2y=6-2\sqrt5\end{cases}\Rightarrow\begin{cases}x=0\\ y=3-\sqrt5\end{cases}\)

<=> xy+5x+3y+15=xy+8x+y+8                 <=> 3x-2y=7           <=>  9x-6y=21 <=> x=3            <=> x=3

      10xy+14x-15y-21=10xy+10x-12y-12            4x-3y=9                  8x-6y=18       8.3-6y=18           y=1

moi hok lop 6 thoi

em moi hoc lop 6 thoi sao lam duoc toan lop 9

Grade 5 students only know how to do

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)