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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
=>\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
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a,\(\left(x-\frac{2}{3}\right),\left(x+\frac{1}{1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2}{3}\\x+\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-1}{4}\end{matrix}\right.\)
b,\(\left(x-\frac{2}{3}\right)\left(2x-\frac{3}{4}\right)=\left(3x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\)
\(\Leftrightarrow2x^2-\frac{3}{4}x-\frac{4}{3}x+\frac{1}{2}=3x^2+2x+\frac{1}{2}x+\frac{1}{3}\)
\(\Leftrightarrow2x^2-\frac{25}{12}x+\frac{1}{2}=3x^2+\frac{5}{2}x+\frac{1}{3}\)
\(\Leftrightarrow24x^2-25x+6=36x^2+30x+4\)
\(\Leftrightarrow24x^2-25x+6-36x^2-30x-4=0\)
\(\Leftrightarrow-12x^2-55x+2=0\)
\(\Leftrightarrow12x^2+55x-2=0\)
\(\Leftrightarrow x=\frac{-55\pm\sqrt{55^2-4.12\left(-2\right)}}{2.12}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3025+96}}{24}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3121}}{24}\)
\(\Leftrightarrow\frac{-55+\sqrt{3121}}{24}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-55+\sqrt{3121}}{24}\\\frac{-55-\sqrt{3121}}{24}\end{matrix}\right.\)
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 ....
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
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
Bài 1:
a) \(\left(2-3x\right)-\left(5x+8\right)=15x\)
\(\Leftrightarrow2-3x-5x-8-15x=0\)
\(\Leftrightarrow-23x-6=0\)
\(\Leftrightarrow x=\frac{-6}{23}\)
Vậy...
b) \(3\left(x-3\right)-2\left(8-x\right)=6\)
\(\Leftrightarrow3x-9-16+2x-6=0\)
\(\Leftrightarrow5x-31=0\)
\(\Leftrightarrow x=\frac{31}{5}\)
Vậy...
c) \(\frac{7-x}{2}-\frac{2x-3}{4}=\frac{x+2}{8}-\frac{-1}{2}\)
\(\Leftrightarrow4\left(7-x\right)-2\left(2x-3\right)=x+2+4\)
\(\Leftrightarrow28-4x-4x+6-x-6=0\)
\(\Leftrightarrow-9x+28=0\)
\(\Leftrightarrow x=\frac{28}{9}\)
Vậy...
d) \(x^2\cdot\left(-4x\right)+3=0\)
\(\Leftrightarrow-4x^3=-3\)
\(\Leftrightarrow x^3=\frac{3}{4}\)
\(\Leftrightarrow x=\sqrt[3]{\frac{3}{4}}\)
Vậy...
a) \(\left(2-3x\right)-\left(5x+8\right)=15x\)
\(\Leftrightarrow2-3x-5x-8=15x\)
\(\Leftrightarrow15x+3x+5x=2-8\)
\(\Leftrightarrow23x=-6\)
\(\Leftrightarrow x=-\frac{6}{23}\)
Vậy : \(x=-\frac{6}{23}\)
b) \(3\left(x-3\right)-2\left(8-x\right)=6\)
\(\Leftrightarrow3x-9-16+2x=6\)
\(\Leftrightarrow5x=6+9+16=41\)
\(\Leftrightarrow x=\frac{41}{5}\)
Vậy : \(x=\frac{41}{5}\)
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
a, x=7/6; b, x=61/9 và=67/9
a, /\(\frac{7}{2}\)-3x/ = 0
\(\Leftrightarrow\)\(\frac{7}{2}\)-3x = 0
3x =\(\frac{7}{2}\)
x =\(\frac{7}{6}\)
Vậy x =\(\frac{7}{6}\)
b, / ( -2\(\frac{2}{3}\))2 - x / = \(\frac{1}{3}\)
/ (\(\frac{-8}{3}\))2 - x / =\(\frac{1}{3}\)
/ \(\frac{64}{9}\)- x / =\(\frac{1}{3}\)
=> \(\frac{64}{9}\)- x = \(\frac{1}{3}\)hoặc\(\frac{-1}{3}\)
Với \(\frac{64}{9}\)- x =\(\frac{1}{3}\)
x = \(\frac{64}{9}\)-\(\frac{1}{3}\)
x = \(\frac{61}{9}\)
Với\(\frac{64}{9}\)- x =\(\frac{-1}{3}\)
x =\(\frac{64}{9}+\frac{1}{3}\)
x =\(\frac{67}{9}\)
Vậy ...
Nhớ line cho mình nhé!