\(f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\)

\(\Rightarrow a-b+c=-3\)

\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\Rightarrow4a+2b+c=-3\)

\(\Rightarrow3a+3b=0\Rightarrow a=-b\)

\(\Rightarrow a^{2019}=-b^{2019}\Rightarrow a^{2019}+b^{2019}=0\)

\(\Rightarrow A=0\)

a)\(M=\frac{2019\times2020-2}{2018+2018\times2020}=\frac{2019\times2020-2}{2018+2018\times2020+2020-2020}=\frac{2019\times2020-2}{\left(2018+1\right)\times2020+2018-2020}=\frac{2019\times2020-2}{2019\times2020-2}=1\\ N=\frac{-2019\times20202020}{20192019\times2020}=\frac{-2019\times10001\times2020}{2019\times10001\times2020}=-1\)

b)\(5\left|x-1\right|=3M-2N=5\\ \left|x-1\right|=1\Rightarrow\hept{\begin{cases}x-1=1\Rightarrow x=2\\x-1=-1\Rightarrow x=0\end{cases}}\)

a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

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

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

\(=\left(x+1\right)^2\)

\(=x^2+2x+1\)

Bài 1:

b: \(\frac{\frac37-\frac{3}{13}+\frac{3}{17}}{\frac57-\frac{5}{13}+\frac{5}{17}}\)

\(=\frac{3\left(\frac17-\frac{1}{13}+\frac{1}{17}\right)}{5\left(\frac17-\frac{1}{13}+\frac{1}{17}\right)}=\frac35\)

c: \(\frac{5^4\cdot4^4\cdot5^4}{5^{10}\cdot4^5}=\frac{5^8\cdot4^4}{5^{10}\cdot4^5}=\frac{1}{5^2\cdot4}=\frac{1}{100}\)

d: \(\left|x-2007\right|+\left|2y-2020\right|^{2021}=0\)

=>x-2007=0 và 2y-2020=0

=>x=2007 và 2y=2020

=>x=2007 và y=1010

Bài 2:

b: \(A=\frac{1}{2^2}+\frac{1}{2^4}+\cdots+\frac{1}{2^{100}}\)

=>\(4A=1+\frac{1}{2^2}+\cdots+\frac{1}{2^{98}}\)

=>4A-A=\(1+\frac{1}{2^2}+\cdots+\frac{1}{2^{98}}-\frac{1}{2^2}-\frac{1}{2^4}-\cdots-\frac{1}{2^{100}}\)

=>3A=\(1-\frac{1}{2^{100}}\)

=>3A<1

=>A<1/3

c: \(H=\left|x-3\right|+\left|4+x\right|\)

\(=\left|x+4\right|+\left|3-x\right|\ge\left|x+4+3-x\right|=7\forall x\)

Dấu '=' xảy ra khi (x-3)(x+4)<=0

=>-4<=x<=3

1,

a,\(2020-\left(249+2020\right)+\left(249-573\right)\)

\(=2020-249-2020+249-573\)

\(=-573\)

b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)

\(=257+1-18-257\)

\(=1-18=-17\)

\(c,25.\left(85-47\right)-85.\left(47+25\right)\)

\(=25.85-25.47-47.85+85.25\)

\(=85.\left(25-47+25\right)-25.47\)

\(=85.3-25.47\)

\(=-920\)

2,

\(a,15-5.\left(x+2\right)=-30\)

\(=>5.\left(x+2\right)=15+30=45\)

\(=>x+2=\frac{45}{5}=9\)

\(=>x=7\)

\(b,\left(x+2\right)^2+5=105\)

\(=>\left(x+2\right)^2=100\)

\(=>\left(x+2\right)^2=10^2\)

\(=>x+2=10\)

\(=>x=8\)

\(c,\left|2x-5\right|-\left(-6\right)=11\)

\(=>\left|2x-5\right|=11-6=5\)

\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)

\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)

Theo đề bài ta có :

\(F\left(x\right)=\left(x-1\right)\cdot Q\left(x\right)-4\) (1)

\(F\left(x\right)=\left(x+2\right)\cdot R\left(x\right)+5\) (2)

Thay \(x=1\) vào (1) ta có :

\(F\left(1\right)=-4\)

\(\Leftrightarrow1+a+b+c=-4\)

\(\Leftrightarrow a+b+c=-5\)

Thay \(x=-2\) vào (2) ta có :

\(F\left(-2\right)=5\)

\(\Leftrightarrow-8+4a-2b+c=5\)

\(\Leftrightarrow4a-2b+c=13\)

Do đó ta có : \(\hept{\begin{cases}a+b+c=-4\\4a-2b+c=13\end{cases}}\)

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