
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.




2x chẵn,1 lẻ nên 2x+1 lẻ . Ta có bảng sau :
2x+1 | -3 | -1 | 1 | 3 |
y-5 | -4 | -12 | 12 | 4 |
2x | -4 | -2 | 0 | 2 |
x | -2 | -1 | 0 | 1 |
y | 1 | -7 | 17 | 9 |
x+y | -1 | -8 | 17 | 10 |
Ta thấy x0+y0 lớn nhất là 17 nên (x0;y0) = (0;17) thỏa mãn (2x+1)(y-5) = 12 với x0+y0 lớn nhất.

Bài 1:
a) x.x.y.y.x.y.x = \(x^4.y^3\)
b) 1000.10.10= 1000. \(10^2\)
c) \(3^{15}:3^5=3^{15-5}=3^{10}\)
d) \(9^8:3=3^{16}:3=3^{15}\)
e) \(125:5^3=\dfrac{125}{5^3}=\dfrac{5^3}{5^3}=1\)
B2:

\(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Leftrightarrow\dfrac{9}{3x}+\dfrac{xy}{3x}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{xy+9}{3x}=\dfrac{5}{6}\Leftrightarrow6\left(xy+9\right)=5\cdot3x\)
\(\Leftrightarrow6xy+54=15x\)\(\Leftrightarrow6xy-15x=-54\)
\(\Leftrightarrow3x\left(2y-5\right)=-54\)
\(\Leftrightarrow x\left(2y-5\right)=-18\)

\(\left(x-9\right)^5\left(5+x\right)^8=0\)
\(\Rightarrow\orbr{\begin{cases}x-9=0\\5+x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=9\\x=-5\end{cases}}}\)
\(x+y=2xy\)
\(\Rightarrow x+y-2xy=0\)
\(\Rightarrow\left(\sqrt{x}-\sqrt{y}\right)^2=0\)
\(\Rightarrow x=y\)
Đề câu 3 có vấn đề nha bn

Câu 4:
a) Ta có: \(\left|-x+8\right|\ge0\)
\(\Rightarrow A=\left|-x+8\right|-21\ge-21\)
Vậy \(MIN_A=-21\) khi x = 8
b) Ta có: \(\left|-x-17\right|+\left|y-36\right|\ge0\)
\(\Rightarrow B=\left|-x-17\right|+\left|y-36\right|+12\ge12\)
Vậy \(MIN_B=12\) khi \(x=-17;y=36\)
c) Ta có: \(-\left|2x-8\right|\le0\)
\(\Rightarrow C=-\left|2x-8\right|-35\le-35\)
Vậy \(MAX_C=-35\) khi \(x=4\)
d) Ta có: \(3\left(3x-12\right)^2\ge0\)
\(\Rightarrow D=3\left(3x-12\right)^2-37\ge-37\)
Vậy \(MIN_D=-37\) khi x = 4
e) Ta có: \(-3\left|2x+50\right|\le0\)
\(\Rightarrow E=-21-3\left|2x+50\right|\le-21\)
Vậy \(MAX_E=-21\) khi x = -25
g) \(\left(x-3\right)^2+\left|x^2-9\right|\ge0\)
\(\Rightarrow G=\left(x-3\right)^2+\left|x^2-9\right|+25\ge25\)
Vậy \(MIN_G=25\) khi x = 3