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\(\dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}>-3\\ \dfrac{x+1}{2953}+\dfrac{x+953}{2001}+\dfrac{x+2950}{4}+3>-3+3\\ \dfrac{x+2954}{2953}+\dfrac{x+2954}{2001}+\dfrac{x+2954}{4}>0\\ \left(x+2954\right)\left(\dfrac{1}{2953}+\dfrac{1}{2001}+\dfrac{1}{4}\right)>0\\ x+2954>0\\ x>-2954\)

C.

\(\left|2x-1\right|=\left|x+1\right|\\ \Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-1+x+1\right)\left(2x-1-x-1\right)=0\\ \Leftrightarrow3x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

vậy...

Tong chieu dai va chieu rong la : 5:2=2,5

Chieu rong la : 2,5 :(3+2)x2=1

Chieu dai la : 2,5-1=1,5

Dien tich la : 1x1,5=1,5 m2 tick dung nha

theo đề bài, ta có:

\(F\left(1\right)=Q\left(-1\right)\\ \Leftrightarrow1+2m+m^2=1-3m+2m^2\\ \Leftrightarrow m^2-5m=0\\ \Leftrightarrow m\left(m-5\right)=0\Rightarrow\left[{}\begin{matrix}m=0\\m=5\end{matrix}\right.\)

vậy F(1)=Q(-1) khi x=0 hoặc 5

396

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\(a.\: 2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\\ =\left(x+y\right)\left(2a^2b+4a^3b\right)\\ =2a^2b\left(x+y\right)\left(1+2a\right)\)

\(b.\:-3a\left(x-y\right)-a^2\left(7-x\right)\\ =a\left(3y-3x-7a+ax\right)\)

ta thấy : 16³-64²=0

do đó kết quả cũng sẽ =0