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đề bài đâu ?

NHưng giải thik về j cơ ???

²2 là thế nào đấy bạn?

2 mủ 2 đấy bn

\(1,\frac{-x}{15}=\frac{10}{25}\)

\(\Leftrightarrow\frac{-x}{15}=\frac{2}{5}\Leftrightarrow\left(-x\right).5=2.15\)

\(\Leftrightarrow\left(-x\right).5=30\)

\(\Leftrightarrow-x=6\Rightarrow x=-6\)

\(2,\frac{x+1}{14}=\frac{10}{35}\)

\(\Leftrightarrow\frac{x+1}{14}=\frac{2}{7}\)

\(\Leftrightarrow\left(x+1\right).7=2.14\)

\(\Leftrightarrow7x+7=28\)

\(\Leftrightarrow7x=21\)

\(\Rightarrow x=3\)

\(3,\frac{\left|x+2\right|}{26}=\frac{12}{39}\)

\(\Leftrightarrow\frac{\left|x+2\right|}{26}=\frac{4}{13}\)

\(\Leftrightarrow\left|x+2\right|.13=4.26\)

\(\Leftrightarrow\left|x+2\right|.13=104\)

\(\Leftrightarrow\left|x+2\right|=8\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=8\\x+2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

Vậy x = 6 hoặc x = -10

\(4,x-\left(1\frac{1}{3}-\frac{2}{3}\right)=2\)

\(\Leftrightarrow x-\left(\frac{4}{3}-\frac{2}{3}\right)=2\)

\(\Leftrightarrow x-\frac{2}{3}=2\)

\(\Leftrightarrow x=2+\frac{2}{3}\)

\(\Leftrightarrow x=\frac{8}{3}\)

\(5,\frac{2}{5}:x-\frac{1}{5}=\frac{3}{5}\)

\(\Leftrightarrow\frac{2}{5}:x=\frac{3}{5}+\frac{1}{5}\)

\(\Leftrightarrow\frac{2}{5}:x=\frac{4}{5}\)

\(\Leftrightarrow x=\frac{2}{5}:\frac{4}{5}\)

\(\Rightarrow x=\frac{1}{2}\)

\(6,\frac{1}{2}.\left(3x-\frac{2}{5}\right)=\left(-3\right)^2\)

\(\Leftrightarrow\frac{1}{2}.\left(3x-\frac{2}{5}\right)=9\)

\(\Leftrightarrow3x-\frac{2}{5}=9:\frac{1}{2}\)

\(\Leftrightarrow3x-\frac{2}{5}=18\)

\(\Leftrightarrow3x=\frac{92}{5}\)

\(\Leftrightarrow x=\frac{92}{5}:3\)

\(\Rightarrow x=\frac{92}{15}\)

\(\frac{-x}{15}=\frac{10}{25}\)

\(\Leftrightarrow25\times\left(-x\right)=10\times15\)

\(-25x=150\)

\(x=150\div\left(-25\right)\)

\(x=-6\)

a: \(\frac{375\cdot833+822}{377\cdot833-844}\)

\(=\frac{375\cdot833+822}{375\cdot833+2\cdot833-844}=\frac{375\cdot833+822}{375\cdot833+822}\)

=1

b: \(\frac{134^2-670}{133^2-532}=\frac{134\left(134-5\right)}{133\left(133-4\right)}=\frac{134\cdot129}{133\cdot129}=\frac{134}{133}=1+\frac{1}{133}\)

\(\frac{7\cdot\left(13\cdot14-26\cdot28\right)}{26\cdot28-52\cdot56}=\frac{7\cdot13\cdot14\cdot\left(1-2\cdot2\right)}{26\cdot28\left(1-2\cdot2\right)}\)

\(=\frac{7\cdot13\cdot14}{26\cdot28}=7\cdot\frac12\cdot\frac12=\frac74=1+\frac34\)

Ta có: \(\frac{1}{133}=\frac{3}{399}<\frac34\)

=>\(\frac{1}{133}+1<\frac34+1\)

=>\(\frac{134^2-670}{133^2-532}<\frac{7\cdot\left(13\cdot14-26\cdot28\right)}{26\cdot28-52\cdot56}\)

c: \(\frac{2019\cdot2020-1}{2019\cdot2020}=1-\frac{1}{2019\cdot2020}\)

\(\frac{2020\cdot2021-1}{2020\cdot2021}=1-\frac{1}{2020\cdot2021}\)

Ta có: \(2019\cdot2020<2020\cdot2021\)

=>\(\frac{1}{2019\cdot2020}>\frac{1}{2020\cdot2021}\)

=>\(-\frac{1}{2019\cdot2020}<-\frac{1}{2020\cdot2021}\)

=>\(-\frac{1}{2019\cdot2020}+1<-\frac{1}{2020\cdot2021}+1\)

=>\(\frac{2019\cdot2020-1}{2019\cdot2020}<\frac{2020\cdot2021-1}{2020\cdot2021}\)

d: \(31^8<32^8=\left(2^5\right)^8=2^{40}\)

\(128^6=\left(2^7\right)^6=2^{42}>2^{40}\)

=>\(31^8<128^6\)

=>\(\frac{1}{31^8}>\frac{1}{128^6}\)

=>\(\left(\frac{1}{31}\right)^8>\left(\frac{1}{128}\right)^6\)

câu này mình tl rùi

tính j

2006 . 125 + \(\dfrac{1000}{126}\) . 2005 - 888 = 265774,6984

Ko hiểu cái đề luôn!