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3) \(\dfrac{3}{4}.x-\dfrac{5}{3}.x=\dfrac{7}{12}\)
\(\left(\dfrac{3}{4}-\dfrac{5}{3}\right).x=\dfrac{7}{12}\)
\(-\dfrac{11}{12}.x=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}:\left(-\dfrac{11}{12}\right)\)
\(x=-\dfrac{7}{11}\)
Ta có:
\(\text{ a)(-14).(-125).(+3).(-8)}\)
\(=\left[\left(-14\right).\left(+3\right)\right].\left[\left(-125\right).\left(-8\right)\right]\)
\(=\left(-42\right).1000\)
\(=-42000\)
\(b)\left(-127\right).57+\left(-127\right).43\)
\(=\left(-127\right).\left(57+43\right)\)
\(=\left(-127\right).100\)
\(=-12700\)
\(c)\left(-13\right).34-87.34\)
\(=34.\left[\left(-13\right)-87\right]\)
\(=34.\left(-100\right)\)
\(=-3400\)
#Mạt Mạt#
mk làm bài 2 trước nhé
\(\frac{x+2}{2}=\frac{72}{x+2}\)
\(=>\left(x+2\right)^2=72.2=144=12^2\)
\(=>x+2=12\)
\(=>x=12-2=10\)
\(1+3+5+7+...+99=\left(x-2\right)^2\)
\(\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)=\left(x-2\right)^2\)
\(100+100+100+...+100=\left(x-2\right)^2\)
\(100\cdot25=\left(x-2\right)^2\)
\(2500=\left(x-2\right)^2\)
\(\pm50^2=\left(x-2\right)^2\)
\(50=x-2\) hoặc \(-50=x-2\)
\(x=50+2\) hoặc \(x=-50+2\)
\(x=52\) hoặc \(x=-48\)
\(\left(1+99\right).49:100=2450\)