Tim x biet
c) (x - 2)(x + 3)>0
d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
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\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=\frac{-32}{27}-\left(-\frac{24}{27}\right)\)
\(\left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\)
\(3x=\frac{-2}{3}+\frac{7}{9}\)
\(3x=\frac{1}{9}\)
\(x=\frac{1}{9}:3\)
\(x=\frac{1}{27}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
a)\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
=\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3^2}{5^2}\)
=\(x+\frac{1}{5}=\frac{3}{5}\)
\(x=\frac{2}{5}\)
b)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\)
=\(x=-\frac{35}{27}\)
a) \(\frac{x}{5}-\frac{x}{6}=\frac{3}{10}\\ \frac{6x}{30}-\frac{5x}{30}=\frac{3\cdot3}{10\cdot3}\\ \frac{x}{30}=\frac{9}{30}\\ \Rightarrow x=9\) Vậy x = 9
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\\ \frac{-32}{27}+\frac{24}{27}=\left(3x-\frac{7}{9}\right)^3\\ \left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}\\ \left(3x-\frac{7}{9}\right)^3=\left(\frac{-2}{3}\right)^3\\ \Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\\ 3x=\frac{-2}{3}+\frac{7}{9}\\ 3x=\frac{-6}{9}+\frac{7}{9}\\ 3x=\frac{1}{9}\\ x=\frac{1}{9}:3\\ x=\frac{1}{9\cdot3}\\ x=\frac{1}{27}\)Vậy \(x=\frac{1}{27}\)
a,(5/8/17+-4/17):x+33/182=4/11
=5/4/17:x+33/182=4/11
5/4/17:x=4/11-33/182
5/4/17:x=365/2002
x=5/4/17:365/2002
x=28/4438/6205
b,-1/5/27-(3x-7/9)^3=-24/27
(3x-7/9)^3=-1/5/27--24/27
(3x-7/9)^3=-8/27
(3x-7/9)^3=(-2/3)^3
3x-7/9=-2/3
3x=-2/3+7/9
3x=1/9
x=1/9:3
x=1/27
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
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!!!!