B = \(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=\frac{-240+270+60-144}{360}=\frac{-54}{360}=-0,15\)

giup minh voi cac ban

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(=0\)

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)

a. Tại x=\(\frac{-1}{2}\), ta có:

 \(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)

b. Ta có:

 \(x^2+4x+3=0\)

\(\Rightarrow x^2+x+3x+3=0\)

\(\Rightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)

\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)

Vậy \(x=-1;x=-3\)

D = \(\frac{9}{4}+\frac{1}{16}-\frac{1}{8}\)

D = \(\frac{35}{16}\)

\(D=\left(\frac{3}{2}\right)^2+\left(\frac{1}{4}\right)^2-\left(\frac{1}{2}\right)^3\)

\(D=\frac{9}{4}+\frac{1}{16}-\frac{1}{8}\)

\(D=\frac{37}{16}-\frac{1}{8}\)

\(D=\frac{35}{16}.\)

Chúc bạn học tốt!

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\left(-\sqrt{7}\right)+\left(-\sqrt{5}\right)\right)\cdot\frac{\sqrt{7}-\sqrt{7}}{1}\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\cdot\frac{\sqrt{7}-\sqrt{5}}{1}\)

\(=\frac{-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{1}\)

\(=\frac{-\left(7-5\right)}{1}=-2\)

\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

\(=\frac{99}{100}\)

99/100 đúng đó nhơ minh nhé hi hi