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a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
Bài 1:
a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)
hay \(x=\frac{77}{72}\)
Vậy: \(x=\frac{77}{72}\)
b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)
\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)
hay \(x=-\frac{46}{35}\)
Vậy: \(x=-\frac{46}{35}\)
c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)
\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)
hay \(x=\frac{2}{3}\)
Vậy: \(x=\frac{2}{3}\)
d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)
\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)
hay \(x=\frac{25}{42}\)
Vậy: \(x=\frac{25}{42}\)
e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)
\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)
\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)
hay \(x=-\frac{4}{5}\)
Vậy: \(x=-\frac{4}{5}\)
f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)
\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)
\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)
Vậy: \(x=-\frac{2}{3}\)
g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)
mk ko chép lại đề nhé bn
b,
=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-\frac{14}{5}\right|\)
=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\) \(\Rightarrow\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{7}{3}\end{cases}}}\)
c,\(\Rightarrow\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)
=> \(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\right)=0\)
=>\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)
=.\(\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
Do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)=> x-2014=0
=> x=2014
d,\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x-1}.\left[1-\left(x-7\right)^{x+12}\right]=0\)
=> \(\orbr{\begin{cases}\left(x-7\right)^{x-1}=0\\1-\left(x-7\right)^{x+12}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x+12}=0\end{cases}}\)
=>x=7 hoặc x-7=1 hoặc x+12=0
=> x=7 hoặc x=8 hoặc x=-12
Vậy x=7, x=8, x=-12
k,3x+x2=0
=> x(3+x)=0
=>\(\orbr{\begin{cases}x=0\\3+x=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
m, x2-2x-3(x-2)=0
=> x(x-2)-3(x-2)=0
=> (x-3)(x-2)=0
=>\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
*****Chúc bạn học giỏi*****
a) \(2x\left(x-\frac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy \(x=0;x=\frac{1}{7}\)
b) \(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\\ \left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\\ \left(\frac{5}{10}+\frac{6}{10}\right)x=\frac{-33}{25}\\ \frac{11}{10}x=\frac{-33}{25}\\ x=\frac{-33}{25}:\frac{11}{10}\\ x=\frac{-33.10}{25.11}\\ x=\frac{-6}{5}\)
Vậy x = \(\frac{-6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}:x=\frac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4.3}{9.2}=\frac{2}{3}\\x=\frac{-3}{7}:\frac{-1}{2}=\frac{-3.2}{7.\left(-1\right)}=\frac{6}{7}\end{matrix}\right.\)
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)
b) \(\frac{1}{2}x+\frac{3}{5}x=-\frac{33}{25}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)
\(\Rightarrow x\frac{11}{10}=-\frac{33}{25}\)
\(\Rightarrow x=\left(-\frac{33}{25}\right):\frac{11}{10}=-\frac{33}{25}.\frac{10}{11}=-\frac{6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=0+\frac{4}{9}=\frac{4}{9}\\-\frac{3}{7}:x=0-\frac{1}{2}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4}{9}.\frac{3}{2}=\frac{2}{3}\\x=\left(-\frac{3}{7}\right):\frac{-1}{2}=\left(-\frac{3}{7}\right).\left(-2\right)=\frac{6}{7}\end{matrix}\right.\)
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow2x-2.\frac{1}{7}=0\)
\(2x-\frac{2}{7}=0\)
=> \(2x=\frac{2}{7}\)
=> x=\(\frac{1}{7}\)
b) (x-9)(\(x+\frac{3}{5}\))=0
\(\Rightarrow\orbr{\begin{cases}x-9=0\\x+\frac{3}{5}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-3}{5}\end{cases}}\)
Vậy x=0 hoặc x=-3/5
c) \(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-4}{7}-2x=0\\x-\frac{5}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{5}{4}\end{cases}}\)
Vậy x=-2/7 hoặc x=5/4
a, => x.(x-1/7) = 0:2 = 0
=> x=0 hoặc x-1/7=0
=> x=0 hoặc x=1/7
Vậy x thuộc {0;1/7}
b, => x-9=0 hoặc x+3/5=0
=> x=9 hoặc x=-3/5
Vậy x thuộc {-3/5;9}
c, => -4/7-2x=0 hoặc x-5/4=0
=> x=-2/7 hoặc x=5/4
Vậy x thuộc {-2/7;5/4}
Tk mk nha