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\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
Lời giải:
a)
\(-3\frac{5}{8}+\left(-\frac{3}{8}+\frac{9}{4}\right)\)
\(=-\frac{29}{8}+\left(-\frac{3}{8}+\frac{18}{8}\right)\)
\(=-\frac{29}{8}+\frac{15}{8}=-\frac{14}{8}=-\frac{7}{4}\)
b) \(\frac{\left(-9\right)\cdot11+32\cdot\left(-9\right)}{\left(-43\right)\cdot15+12\cdot\left(-43\right)}=\frac{\left(-9\right)\left(11+32\right)}{\left(-43\right)\left(15+12\right)}=\frac{\left(-9\right)\cdot43}{\left(-43\right)\cdot27}=\frac{\left(-1\right)\cdot1}{\left(-1\right)\cdot3}=\frac{1}{3}\)
c) Thay \(x=\frac{2011}{2012}\)vào biểu thức \(x\cdot\frac{1}{3}+2x\cdot\frac{3}{6}-3x\cdot\frac{4}{9}\)ta có :
\(\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{3}{6}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{2012}\cdot\frac{1}{3}+2\cdot\frac{2011}{2012}\cdot\frac{1}{2}-3\cdot\frac{2011}{2012}\cdot\frac{4}{9}\)
\(=\frac{2011}{6036}+\frac{2011}{2012}-\frac{2011}{1509}\)
\(=\frac{2011}{6036}+\frac{6033}{6036}-\frac{8044}{6036}=\frac{2011+6033-8044}{6036}=0\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
b)
\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(21.\frac{1}{2}\right)\)
\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}.\frac{76}{45}\right):\frac{21}{2}\)
\(\Rightarrow2-7< x< \left(1+\frac{38}{5}\right):\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{43}{5}:\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{86}{105}\)
Vì \(x\in Z\left(gt\right)\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}.\)
Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}.\)
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)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
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
tự làm ak?