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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a: =>2x-1=4 hoặc 2x-1=-4
=>2x=5 hoặc 2x=-3
=>x=5/2 hoặc x=-3/2
d: =>x=|2|=2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-y=0\end{matrix}\right.\Rightarrow x=y=1\)
b. + Vì \(|6-2x|\ge0\)\(\forall x\)
\(\Rightarrow\)\(|6-2x|-5\ge0-5\)\(\forall x\)
\(\Rightarrow\)B\(\ge\)-5 \(\forall x\)
Vậy GTNN của B= -5 \(\Leftrightarrow\)6-2x=0
\(\Leftrightarrow\)2x=6
\(\Leftrightarrow\)x=3
+ Vì -\(|6-2x|\le0\forall x\)
\(\Rightarrow\)\(|6-2x|-5\le0+5\forall x\)
\(\Rightarrow B\le5\forall x\)
Vậy GTLN của B= 5 \(\Leftrightarrow6-2x=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
c,+ Vì \(|x+1|\ge0\forall x\)
\(\Rightarrow\)\(3-|x+1|\ge3-0\forall x\)
\(\Rightarrow C\ge3\forall x\)
Vậy GTNN của C=3 \(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
+ Vì \(-|x+1|\le0\forall x\)
\(\Rightarrow3-|x+1|\le3+0\forall x\)
\(\Rightarrow C\le3\forall x\)
Vậy GTLN của \(C=3\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Mình chỉ làm vậy thôi nhé!
\(a,x.\frac{-3}{7}=\frac{4}{21}\)
\(x=\frac{4}{21}:\frac{-3}{7}\)
\(x=\frac{-4}{9}\)
\(b,\frac{-4}{7}:x=\frac{2}{5}\)
\(x=\frac{-4}{7}:\frac{2}{5}\)
\(x=\frac{-10}{7}\)
\(c,x+\frac{1}{12}=\frac{-3}{8}\)
\(x=\frac{-3}{8}-\frac{1}{12}\)
\(x=\frac{-11}{24}\)
\(d,\frac{2}{15}-x=\frac{-3}{10}\)
\(x=\frac{2}{15}+\frac{3}{10}\)
\(x=\frac{13}{30}\)
\(e,-x+\frac{4}{5}=\frac{1}{2}\)
\(-x=\frac{-3}{10}\)
\(x=\frac{3}{10}\)
\(f,\frac{3}{4}.\left(x+1\right)-\frac{1}{2}=\frac{3}{7}\)
\(\frac{3}{4}.\left(x+1\right)=\frac{13}{14}\)
\(x+1=\frac{26}{21}\)
\(x=\frac{5}{21}\)
\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
\(\frac{-3}{2}-2x=\frac{-11}{4}\)
\(2x=\frac{-3}{2}+\frac{11}{4}\)
\(2x=\frac{-17}{4}\)
\(x=\frac{-17}{8}\)
\(h,-x+\frac{4}{5}=\frac{1}{2}\)
\(-x=\frac{-3}{10}\)
\(x=\frac{3}{10}\)
chúc bạn học tốt !!!
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
e;E = \(\dfrac{x+3}{x-7}\) (\(x\ne\) 7)
E \(\in\) Z ⇔ \(x+3\) ⋮ \(x\) - 7
⇒ \(x-7\) + 10 ⋮ \(x-7\)
⇒ 10 ⋮ \(x-7\)
\(x-7\) \(\in\) Ư(10) = {- 10; - 5; - 2; - 1; 1; 2; 5; 10}
Lập bảng ta có:
Theo bảng trên ta có: \(x\) \(\in\) {- 3; 2; 6; 8; 9; 12; 17}
Vậy \(x\) \(\in\) {-3; 2; 5; 6; 8; 9; 12; 17}
Em cần làm gì với biểu thức này.