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\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
a)\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b)(2x-3)(6-2x)=0
=>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
c)\(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)
\(x:\frac{3}{4}=-\frac{11}{12}\)
\(x=-\frac{11}{16}\)
d)\(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)
\(-\frac{2}{3}x+1=\frac{3}{2}\)
\(-\frac{2}{3}x=\frac{1}{2}\)
\(x=-\frac{3}{4}\)
\(60\%x+\frac{2}{3}x=\frac{1}{3}.6\frac{1}{3}\)
\(\frac{3}{5}x+\frac{2}{3}x=\frac{19}{9}\)
\(\frac{19}{15}x=\frac{19}{9}\)
\(x=\frac{5}{3}\)
a, \(x^2+5< 25\)
\(x^2< 20\)
\(x=1;2;3;4\)
b, \(\frac{10x+8}{10x-15}\in Z\)
\(\left(10x+8\right)-\left(10x-15\right)⋮10x-15\)
\(10x+8-10x+15⋮10x-15\)
\(23⋮10x-15\)
=>\(10x-15\inƯ_{23}=\left\{-1;1;23;-23\right\}\)
\(TH1:10x-15=-1\) \(TH2:10x-15=1\) \(TH3:10x-15=23\)
\(x=\frac{14}{10}\notin Z\) \(x=\frac{16}{10}\notin Z\) \(x=\frac{38}{10}\notin Z\)
\(TH4:10x-15=-23\)
\(x=\frac{8}{10}\notin Z\)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
a, Vì \(\left|3x-6\right|\ge0\) với mọi x
\(\left(x+2\right)^2\ge0\) với mọi x
=> \(\left|3x-6\right|+\left(x+2\right)^2\ge0\)
mà \(\left|3x-6\right|+\left(x+2\right)^2=0\)
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}3x-6=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)
a) /3x-6/+(x+2)^2=0
vì 3x-6 lớn hơn hoặc bằng 0 Với mọi x thuộc Z
(x+2)^2 lớn hơn hoặc bằng 0 Với mọi x thuộc Z
nên /3x-6/+(x+2)^2=0
khi 3x-6=0 suy ra x=2
(x+2)^2=0 suy ra x=-2
vậy x=2 hoặc x=-2
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
sao mà nó dễ zữ vậy bạn
a/
\(\Leftrightarrow\left(2x+\frac{3}{2}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{2}\right)^2=\left(\frac{3}{5}\right)^2hay\left(2x+\frac{3}{2}\right)=\left(-\frac{3}{5}\right)^2\)
\(\Leftrightarrow2x+\frac{3}{2}=\frac{3}{5}hay2x+\frac{3}{2}=-\frac{3}{5}\)
Rồi bạn giải cả 2 trường hợp + kết luận
b/
\(\Leftrightarrow2!x!=\frac{7}{4}\)
\(\Leftrightarrow!x!=\frac{7}{8}\)
\(\Leftrightarrow x=\frac{7}{8}hayx=-\frac{7}{8}\)
c/ \(\Leftrightarrow\hept{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=3\\2x=6\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}}}\)