ai làm đc gửi card 10k vt gồm 1tick nua
giải pt\(\hept{\begin{cases}x^2-2y^2=2x+y\\y^2-2x^2=2y+x\end{cases}}\)
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bàn tròn là bàn ko méo bàn ko méo là mèo ko bán
nha mnk kb cho
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 cục mụn
999999999999999999999999999999999999999999999999999999999999999999999 cục mụn
(2x2-x.)2-9+(2x2-x-3)=0
(2x2-x-3)(2x2-x+3)+(2x2-x-3.)=0
(2x2-x-3.)(2x2-x+4)=0
=> 2x2-x-3=0 (do. 2x2-x+4.>0)
=> 2x2-2-(x+1.)=0
=>2(x+1)(x-1)-(x+1)=0
=> (x+1)(2x-3)=0
=> x1=-1 ; x2=3/2
a)1+2+3+4+5+6+7+8+9=45
b)5+5=10
c)1+1=2
Lấy trên (bỏ ) trừ dưới nhé
\(3\left(x^2-y^2\right)=x-y\Leftrightarrow\left(x-y\right)\left[3\left(x+y\right)-1\right]=0\\ \)
\(\hept{\begin{cases}\left(x-y\right)=0\\\left(x+y\right)=1\end{cases}}\)thế vào @ALI \(\hept{\begin{cases}\left(x+y\right)\left(0-3\right)=0\\\left(1.\right)\left(x-y-3\right)=0\end{cases}}\)\(\hept{\begin{cases}x=-y\\x=y+3\end{cases}}\) cho xin 2 k vt
Lấy trên cộng dưới ta được
x2 - y2 = 3x + 3y
<=> (x + y) (x - y - 3) = 0
<=> x = - y hoặc x = y + 3
Thế ngược lại là giải ra