Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A

Ta có:

\(A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}\) và \(B=\frac{1}{7^{2019}}\)

Ta xét 2 trường hợp:

\(TH1:\frac{4-7^{2020}}{7^{2020}}=\frac{-7^{2020}+4}{7^{2020}}=-1+\frac{4}{7^{2020}}\)

\(TH2:\frac{5+7^{2021}}{7^{2021}}=1+\frac{5}{7^{2021}}\)

\(\Rightarrow\left(-1+\frac{4}{7^{2020}}\right)+\left(1+\frac{5}{7^{2021}}\right)\)

\(\Rightarrow\frac{4}{7^{2020}}+\frac{5}{7^{2021}}\)

\(Do:\)

\(\frac{4}{7^{2020}}>\frac{1}{7^{2019}}\)

\(\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

Nên:\(\frac{4}{7^{2020}}+\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

\(\Rightarrow A>B\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu để của bạn hơn nhé.

Lời giải:

$14.7^{2021}=35.7^{2021}-3.49^x$

$2.7^{2022}=5.7^{2022}-3.7^{2x}$

$3.7^{2x}=5.7^{2022}-2.7^{2022}=7^{2022}(5-2)=3.7^{2022}$

$\Rightarrow 7^{2x}=7^{2022}$

$\Rightarrow 2x=2022$

$\Rightarrow x=2022:2=1011$

tl nhanh giúp mình nha

\(a,\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2021\end{matrix}\right.\\ c,\Leftrightarrow4x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ d,\Leftrightarrow\left(3x+7-x-1\right)\left(3x+7+x+1\right)=0\\ \Leftrightarrow\left(2x+6\right)\left(4x+8\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

\(a)\dfrac{7}{8}=\dfrac{7\times9}{8\times9}=\dfrac{63}{72}\)

\(\dfrac{3}{9}=\dfrac{3\times8}{9\times8}=\dfrac{24}{72}\)

Do : \(\dfrac{63}{72}>\dfrac{24}{72}\) nên \(\dfrac{7}{8}>\dfrac{3}{9}\)

Không thì bạn có thể rút gọn 3/9 đi làm cho nó gọn ạ.

\(b)\) Ta thấy : \(\dfrac{2023}{2021}>1\) ( vì tử lớn hơn mẫu )

                   \(\dfrac{2021}{2022}< 1\) ( vì tử bé hơn mẫu )

Do đó : \(\dfrac{2023}{2021}>\dfrac{2021}{2022}\)

\(c)\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)

\(\dfrac{6}{7}=\dfrac{6\times6}{7\times6}=\dfrac{36}{42}\)

Do : \(\dfrac{36}{42}>\dfrac{35}{42}\)  nên \(\dfrac{6}{7}>\dfrac{5}{6}\)

không câu SP nhé

guihopijp['j'OPHGPG