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\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
a, 1/2x +3/5(x-2)=3
1/2x +3/5.x -3/5.2=3
(1/2+3/5).x-6/5=3
11/10.x=21/5
x=42/11
b,| 1/2x-3/2|= 1/2
=> 1/2x-3/2=1/2
hoac 1/2x-3/2=-1/2
=> 1/2x=2
hoac 1/2x=1
=> x=4 hoac 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
a) 1/3x + 2/5x - 2/5 = 0
=> x = 0,54
b) 12n - 4n^2 - 18 + 6n +0
<=> -4n^2 + 6n - 18 = 0
<=> (-4n)^2 + 6n + 12n - 18 +0
<=> - 2n (2n-3 ) + 6 ( 2n - 3 ) = 0
,<=> ( 6 - 2n ) ( 2n -3 )=0
<=> 6 - 2n = 0 => n +3 / 2n-3 =0 => n = 3/2
a, \(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=0+\frac{1}{16}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2=\left(\frac{-1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{3}=\frac{-1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{11}{12}\\\frac{1}{x}=\frac{5}{12}\end{cases}\Rightarrow\orbr{\begin{cases}11x=12\\5x=12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}}\)
b, \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
Đặt S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)
2S = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}\)
2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2S = \(\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
S = \(\frac{22}{45}:2=\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}\Rightarrow x=\frac{23}{11}\)
a/ (1/x -2/3)2=1/16=(1/4)2
Có 2 trường hợp:
+/ 1/x -2/3= - 1/4
<=> 1/x =2/3 -1/4 = 5/12
=> x1=12/5
+/ 1/x - 2/3 =1/4
<=> 1/x = 2/3 +1/4= 11/12
=> x2=12/11
b/ Ta có:
2/(1.2.3)=1/(1.2) - 1/2.3 ; 2/(2.3.4)=1/2.3 -1/3.4 ; ...; 2/(8.9.10)=1/8.9 -1/9.10
=> (1/1.2.3 + 1/2.3.4 +...+1/8.9.10)=23/45
<=> (1/1.2 -1/2.3 +1/2.3 -1/3.4 +...+1/8.9-1/9.10).x/2=23/45
<=> (1/1.2 -1/9.10).x/2 =23/45
<=> x.11/45=23/45
=> x=23/11
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
Ta có 2 trường hợp :
TH1 : \(\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{12}{11}\)
TH2 : \(\frac{1}{x}-\frac{2}{3}=-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{x}=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{12}{5}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\) <=> \(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
=> \(|\frac{1}{x}-\frac{2}{3}|=\frac{1}{4}\)=> \(\frac{1}{x}-\frac{2}{3}=\pm\frac{1}{4}\)
+/ TH1: \(\frac{1}{x}=\frac{2}{3}+\frac{1}{4}=\frac{11}{12}=>x=\frac{12}{11}\)
+/ TH2: \(\frac{1}{x}=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}=>x=\frac{12}{5}\)