1: tim x:
c : \(\frac{1-x}{3}\) =\(\frac{27}{1-x}\)
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17 tháng 1 2021
Đk: x \(\ne\)0; x \(\ne\)\(\pm\)3
Ta có: A = \(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{x^2+3\left(3-x\right)}{3\left(x+3\right)\left(3-x\right)}\)
A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{3\left(3-x\right)\left(x+3\right)}{x^2-3x+9}\)
A = \(\frac{-\left(x+3\right)}{x}\)
Để A < -1 <=> \(-\frac{\left(x+3\right)}{x}< -1\) <=> \(\frac{-x-3}{x}+1< 0\)
<=> \(\frac{-x-3+x}{x}< 0\) <=> \(-\frac{3}{x}< 0\)
Do -3 <0 => x> 0
Vậy Để A < -1 <=> x > 0 và x khác 3
20 tháng 6 2020
Bài làm:
c) \(\left(x-2\right)\left(x+3\right)>0\)
Ta xét 2 trường hợp sau:
+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)
+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)
Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)
d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{27}\)
Vậy \(x=\frac{1}{27}\)
Học tốt!!!!
2 tháng 8 2016
a) \(\frac{1}{9}=\frac{x}{27}\)
\(\Rightarrow x=\frac{1}{9}\cdot27\)
\(\Rightarrow x=3\)
b) \(\frac{4}{x}=\frac{8}{6}\)
\(\Rightarrow x=4:\frac{8}{6}\)
\(\Rightarrow x=3\)
c) \(\frac{x}{3}-\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\Rightarrow x=\frac{7}{10}\cdot3\)
\(\Rightarrow x=\frac{21}{10}=2,1\)
CL
2 tháng 8 2016
\(a,\frac{1}{9}\)=\(\frac{3}{27}\)
\(b,\frac{4}{3}\)=\(\frac{8}{6}\)
\(c,\frac{x}{3}\)-\(\frac{1}{2}=\frac{1}{5}\)
\(\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(\)
9 tháng 4 2022
a: ĐKXĐ: x<>-3
b: \(Q=\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\dfrac{1}{x+3}\right)\cdot\dfrac{x+3}{x^2-1}\)
\(=\dfrac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\cdot\dfrac{x+3}{x^2-1}\)
\(=\dfrac{2x^2-2}{x^2-1}\cdot\dfrac{1}{x^2-3x+9}=\dfrac{2}{x^2-3x+9}\)
MT
10 tháng 6 2020
c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}.\frac{4}{3}\)
\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\).
d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)
\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).
30 tháng 1 2022
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
31 tháng 7 2016
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{-6}=\frac{y}{4}=\frac{z}{-\frac{1}{2}}\)\(=\frac{4x+y-2z}{-6.4+4-\left(-\frac{1}{2}.2\right)}\)\(=\frac{27}{-24+4+1}=\frac{27}{-19}\)
\(\frac{x}{-6}=\frac{-27}{19}\)\(=>x=\frac{\left(-27\right).\left(-6\right)}{19}=\frac{162}{19}\)
31 tháng 7 2016
Ta có:
\(\frac{x}{-6}\)=\(\frac{4x}{-24}\)
\(\frac{z}{\frac{-1}{2}}\)=\(\frac{2z}{-1}\)
Theo tính chất của dãy tỉ số = nhau:
\(\frac{4x}{-24}\)= \(\frac{y}{4}\)=\(\frac{2z}{-1}\)= \(\frac{4x+y-2z}{-24+4-\left(-1\right)}\)= \(\frac{-27}{19}\)
\(\frac{x}{-6}\)= \(\frac{-27}{19}\)nên x= \(\frac{162}{19}\)= \(8\frac{10}{19}\)
\(\frac{y}{4}\)= \(\frac{-27}{19}\)nên y= \(\frac{-108}{19}\)=\(-5\frac{13}{19}\)
\(\frac{z}{-\frac{1}{2}}\)= \(\frac{-27}{19}\)nên z= \(\frac{27}{38}\)
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
c) Giải:
\(\frac{1-x}{3}=\frac{27}{1-x}\\ \Leftrightarrow\left(1-x\right)\left(1-x\right)=27.3\\ \Rightarrow\left(1-x\right)^2=81\\ \Rightarrow\left(1-x\right)^2=\pm9^2\\ 1-x=\pm9\)
+) 1-x=9
x=1-9
x=-8
+) 1-x=-9
x=1-(-9)
x=10
Vậy \(x\in\left\{-8;10\right\}\)
Chúc bạn học tốt!
\(\Rightarrow\left(x-1\right)\left(1-x\right)=3.27\)
\(\Rightarrow\left(x-1\right)\left(-1\right)\left(x-1\right)=81\)
\(\Rightarrow\left(x-1\right)^2=-81\)
Mặt khác \(\left(x-1\right)^2\ge0\) với mọi x
=> \(x\in\varnothing\)