a) \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}\)

\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2n\left(2n+2\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n+2}\right)\)

b) \(A=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)

\(< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n\left(2n+2\right)}\)

\(< \frac{1}{4}\)

\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)

\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow P=3^2-2Q=9-2=7\)

\(\left(5x-4y\right)^2-\left(6x+4y\right).\left(5x-4y\right)+\left(3x+2y\right)^2\)

\(=\left(5x-4y\right)^2-2.\left(5x-4y\right).\left(3x+2y\right)+\left(3x+2y\right)^2\)

\(=\left(5x-4y-3x-2y\right)^2\)

\(=\left(2x-6y\right)^2\)

\(=4x^2-24xy+36y^2\)

Cho tam giác ABC có D, E lần lượt là trung điểm của AB và AC.

Biết BC = 12cm. Độ dài DE bằng:

A.12cm 

B.6cm

C.8cm

D.7cm

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

\(\Leftrightarrow a^2-2a+1+b^2-2b+1=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\end{cases}}\Leftrightarrow a=b=1\)

\(A=a^{2022}+b^{2022}=2\)

\(5x-15x^2=2\)

\(\Leftrightarrow15x^2-5x+2=0\)

Bấm máy tính ( hoặc có thể tính bằng cách phân tích thành nhân tử hay biệt thức )

=> Phương Trình vô nghiệm

\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)

\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow P=3^2-2Q=9-2=7\)

\(x^4-x^3+x^2-x=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^3+x\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x^2+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow-2\left(2x-7\right)=0\)

\(\Leftrightarrow x=\frac{7}{2}\)

nhầm hà mã emk nhé

꧁༺ＡＧＮＥＳＣＨＡＮＧＧ༻꧂ là s ạ