\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

d. 2x²(x - y) + 2y(y - x)

= 2x²(x - y) - 2y(x - y)

= (2x² - 2y)(x - y)

= 2(x² - y)(x - y)

e. 5a²b(a - 2b) - 2a(2b - a)

= 5a²b(a - 2b) + 2a(a - 2b)

= (5a²b + 2a)(a - 2b)

= a(5ab + 2)(a - 2b)

f. 4x²y(x - y) + 9xy²(x - y)

= (4x²y + 9xy²)(x - y)

= xy(4x + 9y)(x - y)

g. 50x²(x - y)² - 8y²(y - x)²

= 50x²(x² - 2xy + y²) - 8y²(y² - 2xy + x²)

= 50x²(x² - 2xy + y²) - 8y²(x² - 2xy + y²)

= 50x²(x - y)² - 8y²(x - y)²

= (50x² - 8y²)(x - y)²

= 2(25x² - 4y²)(x - y)².

g: Ta có: \(50x^2\left(x-y\right)^2-8y^2\left(y-x\right)^2\)

\(=2\left(x-y\right)^2\left(25x^2-4y^2\right)\)

\(=2\left(x-y\right)^2\left(5x-2y\right)\left(5x+2y\right)\)

TL

XY=60

Học tốt

Sai mik sorry

xem có sai đề ko

không nha bạn

\(\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(2x+2y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left[2\left(x+y\right)+\frac{4}{x+y}\right]^2}{2}\)

\(=8\)

Dấu "=" xảy  ra tại x=y=¹/₂

Có vẻ kết quả  bị sai Huy ơi.

Diệp thay kết quả cuối cùng 8 ------------> 18 nhé!

a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)

\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)

mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)

\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

b) Tương tự câu a, ta có:

\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

c. Tương tự, ta có:

\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...

b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...

c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...

Chân thành cảm ơn các bạn!

vậy bạn tự làm là đk mà

dat x-y=z

suy ra {3z^4+2z^3-5z^2}:z^2

dat nhan tu chung la z^2

=z^2(3z^2+2z-5)

minh chi bt the thoi

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1