\(A=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{46\cdot49}\\ =2\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{46\cdot49}\right)\\=2\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{49}\right)\\ =2\cdot\left(1-\dfrac{1}{49}\right)\\ =\dfrac{96}{49}\)

lm hộc e câu d bài 4

Đề bài bạn yêu cầu gì?

a. CN: nhà văn Hoài Thanh 

VN: khẳng định rằng cái đẹp là có ích 

b. CN1: Tiếng Việt 

VN1: rất giàu thanh điệu 

CN2: lời nói của người Việt Nam chúng ta 

VN2: du dương trầm bổng như bông hoa

\(\dfrac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}=\dfrac{3^6.3^85^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)

\(=\dfrac{3^{14}.5^4-3^{13}.5^4}{2.3^{12}.5^6}=\dfrac{3^{13}.5^4\left(3-1\right)}{2.3^{12}.5^6}\)

\(=\dfrac{3.2}{2.5^2}=\dfrac{3}{25}\)

\(4x=5y\Rightarrow x=\dfrac{5}{4}y\\ \Rightarrow\dfrac{15}{4}y-2y=5\\ \Leftrightarrow\dfrac{7}{4}y=5\\ y=\dfrac{20}{7}\\ \Rightarrow x=\dfrac{25}{7}\)

Bằng 25/7 bạn nhé

\(a=2022.\left|x^2+1\right|+2023\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\left(\left|x^2+1\right|>0,\forall x\right)\)

mà \(\left(x^2+1\right)\ge1,\forall x\)

\(\Rightarrow a=2022.\left(x^2+1\right)+2023\ge2022.1+2023=4045\)

\(\Rightarrow GTNN\left(a\right)=4045\left(x=0\right)\)

GTNN(a) = 4045 khi x = 0

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)