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Ta có: A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{7}-\frac{\sqrt{2}^2}{2^2}\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{7}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-7}{14}\right|+\left|\frac{8+9}{18}\right|\)
= \(\left|\frac{1}{14}\right|+\left|\frac{17}{18}\right|\)
= 1/14 + 17/18 = 64/63
A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}^2}{2^2}\right)\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2.\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{9}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-9}{18}\right|+\left|\frac{4}{9}+\frac{1}{2}\right|\)
= \(\left|-\frac{1}{18}\right|+\left|\frac{8+9}{18}\right|\)
= \(\frac{1}{18}+\frac{17}{18}=1\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)
C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)
C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)
C = \(\frac{-233}{135}\)
D = \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)
D = \(-4.\frac{12}{13}\)
D = \(\frac{-48}{13}\)
E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)
E = \(5.4-4.3+5-0,3.20\)
E = \(20-12+5-6\)
E = \(8+\left(-1\right)\)
E = \(7\)
F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\)
F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)
F = \(\frac{-11}{12}\)
Chúc cậu hk tốt ~
\(\left|\frac{7}{4}-\frac{2}{3}x\right|-\left(\frac{-5}{6}\right)=1-\sqrt{\left(\frac{1}{2}\right)^3}\)
\(\left|\frac{7}{4}-\frac{2}{3}x\right|+\frac{5}{6}=1-\sqrt{\frac{1}{8}}\)
\(\left|\frac{7}{4}-\frac{2}{3}x\right|+\frac{5}{6}=1-\frac{\sqrt{2}}{4}\)
\(\left|\frac{7}{4}-\frac{2}{3}x\right|+\frac{5}{6}=1-0,4\)
\(\left|\frac{7}{4}-\frac{2}{3}x\right|+\frac{5}{6}=0,6\)
\(\left|\frac{7}{4}-\frac{2}{3}x\right|=0,6-\frac{5}{6}=-0,2\)
GTTĐ của một mọi số đều không âm
Mà lại = -0,2
Nên không có giá trị x nào
Đề xấu ~v .Kiểu này chắc x cũng là nghiệm xấu luôn :v =((
\(\left|\frac{7}{4}-\frac{2}{3}x\right|-\left(-\frac{5}{6}\right)=1-\sqrt{\left(\frac{1}{2}\right)^3}\)
\(\Leftrightarrow\left|\frac{7}{4}-\frac{2}{3}x\right|+\frac{5}{6}=1-\frac{\sqrt{2}}{4}\)\(\Leftrightarrow\hept{\begin{cases}\frac{7}{4}-\frac{2}{3}x+\frac{5}{6}=\frac{2-\sqrt{2}}{2}\\\frac{7}{4}-\frac{2}{3}x+\frac{5}{6}=\frac{-\left(2-\sqrt{2}\right)}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{31}{12}-\frac{2}{3}x=\frac{2-\sqrt{2}}{2}\\\frac{31}{12}-\frac{2}{3}x=\frac{-2+\sqrt{2}}{2}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{2}{3}x=\frac{19+6\sqrt{2}}{12}\\\frac{2}{3}x=\frac{43-6\sqrt{2}}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{19+6\sqrt{2}}{8}\\x=\frac{43-6\sqrt{2}}{8}\end{cases}}}\)