\(A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\\ =x^3-xy-x^3-x^2y+x^2y-xy\\ =-2xy\)

Thay `x=1/2;y=-100` vào A ta có:

\(A=-2\cdot\dfrac{1}{2}\cdot\left(-100\right)=100\)

\(B=\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\\=x^3+3x^2-5x-15-x^3+4x+x^2-4x^2\\ =\left(x^3-x^3\right)+\left(3x^2-4x^2+x^2\right)+\left(-5x+4x\right)-15\\ =-x-15\)

\(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}=1,2\\\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=1,2-8\dfrac{1}{5}\\ \Rightarrow \left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=-7\)

Nhận xét:

\(\left\{{}\begin{matrix}\left|2\dfrac{1}{5}-x\right|\ge0,\forall x\\\left|x-\dfrac{1}{5}\right|\ge0,\forall x\end{matrix}\right.\\ \Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0,\forall x\)
Mà \(-7< 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...

\(\left\{{}\begin{matrix}2x+3y=-xy\\\dfrac{8}{x}-\dfrac{6}{y}=5\end{matrix}\right.\left(x;y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\8y-6x=5xy\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=-5xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=5\left(2x+3y\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\6x-8y=10x+15y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=-xy\\-4x=23y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\cdot\dfrac{-23}{4}y+3y=\dfrac{-23}{4}y\cdot y\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y^2-43y=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y\left(23y-43\right)=0\\x=\dfrac{-23}{4}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=0\left(ktm\right)\\y=\dfrac{43}{23}\left(tm\right)\end{matrix}\right.\\x=\dfrac{-23}{4}\cdot\dfrac{43}{23}=\dfrac{-43}{4}\end{matrix}\right.\)

bài giải

số lớn chia số bé được  3 dư 3, số bé bằng 1/3 số lớn

số lớn là:
33 : ( 3 -1 ) x 3 + 3 = 53,5 

số bé là: 

33: ( 3 - 1 ) = 16,5 

đ/s

Gọi chiều dài là a => chiều rộng = a-4

Theo bài ra: a+(a-4)=80

<=> 2a = 84

=> a = 42

=> chiều rộng = a-4 = 42-4 = 38

Diện tích tấm màn là: 38x42= 1596 m² 

10/9*3/28*9/2*7,5=1890/2520

rút gọn=189/252

học tốt☘

7/5 nha mình ghi nhầm ạ

1 Mrs. Lien suggested organizing an English competition for their students
2 Bob denied taking Susie's calculator
3 David accused Henry of taking his pencil

E xem lại câu 2 với 3 bị lẫn lộn r

\(\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}=9\\\dfrac{1}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{y^2+1}}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}=9\\\dfrac{4}{\sqrt{x^2+1}}-\dfrac{4}{\sqrt{y^2+1}}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8}{\sqrt{x^2+1}}+\dfrac{4}{\sqrt{y^2+1}}+\dfrac{4}{\sqrt{x^2+1}}-\dfrac{4}{\sqrt{y^2+1}}=9+3\\\dfrac{1}{\sqrt{x^2+1}}-\dfrac{1}{\sqrt{y^2+1}}=\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{12}{\sqrt{x^2+1}}=12\\\dfrac{1}{\sqrt{y^2+1}}=1-\dfrac{3}{4}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+1=1\\y^2+1=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2=0\\y^2=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{15}\end{matrix}\right.\)