a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\) 

   8 .x + 1 . x = 990

x . [ 8 +1 ] = 990

x . 9 = 990

x = 990 : 9

x = 110

các bạn giúp mình với mình đang vội.

Đáp án C

LÀM ƠN GIÚP MIK ĐI MÀ, NĂN NỈ CÁC BẠN ĐÓ!!!!!!!!!!!!!!!!!!!!!!

16.79

16,79

Ta có:

\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)

\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)

\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)

\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)

\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Đồng thời:

\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự:

\(b^2+1=\left(a+b\right)\left(b+c\right)\)

\(c^2+1=\left(a+c\right)\left(b+c\right)\)

Từ đó:

\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)

Do vế trái luôn dương nên vế phải dường

\(\Rightarrow2021x>0\)

\(\Rightarrow x>0\)

\(\Rightarrow\) Tất cả các số hạng trong trị tuyệt đối đều dương.

Do đó pt trở thành:

\(\left|x+\dfrac{1}{1.2}\right|+\left|x+\dfrac{1}{2.3}\right|+...+\left|x+\dfrac{1}{2020.2021}\right|=2021x\)

\(\Rightarrow x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+x+\dfrac{1}{3.4}+...+x+\dfrac{1}{2020.2021}=2021x\)

\(\Leftrightarrow2020x+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2020.20210}=2021x\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=2021x-2020x\)

\(\Leftrightarrow x=1-\dfrac{1}{2021}\)

\(\Leftrightarrow x=\dfrac{2020}{2021}\)