a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)

d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)

\(a.=\dfrac{2019}{2020}\times\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\\ =\dfrac{2019}{2020}\times1=\dfrac{2019}{2020}\\ b.=\dfrac{25}{27}\times\left(\dfrac{17}{14}-\dfrac{1}{14}-\dfrac{2}{14}\right)\\ =\dfrac{25}{27}\times1=\dfrac{25}{27}\)

Thanks mik chữ cũng hơi ẩu

A=3(x-4)⁴

Vì (x-4)⁴ ≥0

=>3(x-4)⁴ ≥0

Vậy MinA=0

B=5+2(x-2019)²⁰²⁰

Vì (x-2019)²⁰²⁰ ≥0

=>5+(x-2019)²⁰²⁰ ≥5

Để B đạt Min 

=>x-2019=0

=>x=2019

Vậy MinB=5 <=>x=2019

Trả lời:

\(M=\left(x-2020\right)^4+\left(x+y+1\right)^2+5\)

Ta có: \(\left(x-2020\right)^4\ge0\forall x;\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2+5\ge5\forall x,y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2020=0\\x+y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2020\\y=-2021\end{cases}}}\)

Vậy GTNN của M = 5 khi x = 2020; y = - 2021

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