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\(\frac{131.145+100}{45-132.140}=\frac{132.145-45}{45-132.140}=-1\)
\(\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5+2.49^5}=\frac{7^{11}.7-7^{11}.1}{7^{10}.5+2.7^{10}}=\frac{7^{11}.\left(7-1\right)}{7^{10}.\left(5+2\right)}=\frac{7^{11}.6}{7^{11}}=6\)
\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ = - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(\frac{-1}{4}-\frac{5}{3}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=\left(\frac{-3}{12}-\frac{20}{12}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{18}{33}\right)+\frac{48}{49}\)
\(=\left(\frac{-23}{12}+\frac{15}{12}\right)+\left(\frac{-11}{33}\right)+\frac{48}{49}\)
\(=\frac{-2}{3}+\left(\frac{-1}{3}\right)+\frac{48}{49}\)
\(=-1+\frac{48}{49}\)
\(=-\frac{1}{49}\)
Ùng hộ mk nha ^_-
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
\(M=\frac{131.145+100}{45-132.145}\)
\(=\frac{131-\frac{100}{145}}{\frac{45}{145}-132}\)
\(=\frac{131-\frac{20}{29}}{\frac{9}{29}-132}\)
\(=\frac{131\frac{-20}{29}}{-132\frac{9}{29}}\)