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\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)
\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)
Trường hợp 1:
\(2x-\frac{5}{2}=\frac{29}{3}\)
\(2x=\frac{29}{3}+\frac{5}{2}\)
\(2x=\frac{73}{6}\)
\(x=\frac{73}{6}:2\)
\(x=\frac{73}{12}\)
Trường hợp 2:
\(2x-\frac{5}{2}=\frac{-29}{3}\)
\(2x=\frac{-29}{3}+\frac{5}{2}\)
\(2x=\frac{-43}{6}\)
\(x=\frac{-43}{6}:2\)
\(x=\frac{-43}{12}\)
Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(1-\frac{1}{100}=\frac{99}{100}\)
l\(\frac{3}{4}.x-\frac{1}{2}\)l=\(\frac{1}{4}\)
=>\(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\) hoặc \(\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\)
*nếu \(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\)
=>\(\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}=\frac{3}{4}\)
=>\(x=\frac{3}{4}:\frac{3}{4}=1\)
*nếu \(\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\)
=>\(\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=-\frac{1}{4}+\frac{2}{4}=\frac{1}{4}\)
=>\(x=\frac{1}{4}:\frac{3}{4}=\frac{1}{4}.\frac{4}{3}=\frac{1}{3}\)
vậy \(x\in\left\{\frac{1}{3};1\right\}\)
I Am A Good Boy_Wang Jun Kai Ahihi, M lười ghê ý
gọi các phân số cần tìm là\(\frac{a}{12}\)
ta có \(\frac{-2}{3}\)< \(\frac{a}{12}\)< \(\frac{-1}{4}\)
suy ra \(\frac{-8}{12}\)< \(\frac{a}{12}\)< \(\frac{-4}{12}\)
suy ra -8 < a < -4
vậy A thuộc { -7,-6,-5 }
\(\frac{11.3^{29}-\left(3.3\right)^{15}}{2^2.3^{28}}\)=\(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{3^{28}.2^2}=\frac{3.8}{2^2}=3.2\)=6
\(\frac{99}{98}+\frac{96}{97}+\frac{1}{97.98}=\frac{99}{98}+\frac{96}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}+\frac{96}{97}\right)=1+1=2\)
= 2
mk bấm máy tính
đúng 10000000000000000000000000000000000000000000000000000000000000000000000%
\(\frac{7^2.3}{2.3.5^2}\)
\(=\frac{7^2.1}{2.1.5^2}\)
\(=\frac{7^2}{2.5^2}\)
\(=\frac{49}{2.25}\)
\(=\frac{49}{50}\)
\(\frac{7^2.3}{2.3.5^2}=\frac{7^2.3:3}{2.5^2.3:3}=\frac{7^2}{2.5^2}=\frac{49}{50}\)
Tk nha