Giúp mik vs ạ>>
Giải bất phương trình sau:
+ 3x +5 < 14
+ 1 phần x trừ 1 trừ 3 phần x trừ 2 bằng âm 1 phần ( x-1)(x-2)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{3}{2}-\frac{5}{6}:x=\frac{5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{3}{2}-\frac{5}{6}:x=\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{3}{2}-\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{41}{30}\)
\(\Leftrightarrow x=\frac{5}{6}:\frac{41}{30}\)
\(\Leftrightarrow x=\frac{25}{41}\)
b) \(x-\frac{6}{7}.\frac{14}{8}=\frac{1}{2}-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{3}{2}\)
\(\Leftrightarrow x=\frac{8}{5}\)
c) \(x:\frac{6}{5}+\frac{2}{3}=\frac{7}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{7}{3}-\frac{2}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}.\frac{6}{5}\)
\(\Leftrightarrow x=2\)
ĐKXĐ ; \(x\ne\pm1\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)
\(\Leftrightarrow-x^2+4x-3=0\)
\(\Leftrightarrow-x^2+3x+x-3=0\)
\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)
=> X = 3
Vậy ..
a) 3-4x\(\ge\)11
\(4x\le3-11=-8\)
\(x\le-2\)
( câu b bn ghi rõ đề bài đc ko ?)
a) \(3-2x>4\)
\(\Leftrightarrow-2x>1\)
\(\Leftrightarrow x< \frac{-1}{2}\)
b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)
\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow-4x-13-18x+54=x^2-9\)
\(\Leftrightarrow x^2+22x-50=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)
\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )
Vậy....
\(a,\)\(3-2x>4\)
\(\Rightarrow-2x>1\)
\(\Rightarrow x< \frac{-1}{2}\)
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
3x+5<14
<=>3x<9
<=>x<3
vậy S=(x\(\in\)R/x<3)