Tim x biet, \(\left(x-2\right)\times\left(x+\frac{2}{3}\right)>0\)
Những câu hỏi liên quan
tim x biet \((x^2-20)\times\left(x^2-15\right)\left(x^2-10\right)\left(x^2-5\right)< 0\)
Bai 1:a)Tim x biet\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2009}{2011}\)
b)\(\left(x-1\right)\times f\left(x\right)=\left(x+4\right)\times f\left(x\right)\)voi moi x
Bai 2;Tim x;y;z biet a)\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\) b)\(\frac{2x+1}{5}=\frac{3y-z}{7}=\frac{2x+3y-1}{6x}\)
tim x biet
\(\left(x-\frac{1}{3}\right).\left(y-\frac{1}{2}\right).\left(z-5\right)=0\)
và x+2=y+1=z+3
\(\left(x-\frac{1}{3}\right)\left(y-\frac{1}{2}\right)\left(z-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{2}\\z=5\end{cases}}\)
Vì \(z+3=y+1\Rightarrow y=7\)
Lại có \(y+1=x+2\Rightarrow x=8-2=6\)
Vậy x = 6 ; y = 7 ; z = 5
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bai 1: Tim x biet
\(\hept{\begin{cases}x-y=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\)
bai 2: Tim x, y biet:
x+\(\left(-\frac{31}{12}\right)^2\)=\(\left(\frac{49}{12}\right)^2\)-x=y2
Bai 9: Tim x,y,z biet:
(x-1)2+(x+y)2+(xy-z)2=0
a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)
hôm sau mik giải tip cho
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Tim x biet
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(\right)x+10}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Tim x,y,z biet:
\(x+1=y+2=z+3và\left(x-\frac{1}{5}\right)\left(y+\frac{1}{3}\right)\left(z-6\right)=0\)
Tim x biet
a/ \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\left(\frac{x^2-4}{x^2-1}\right)=0\)
b/ \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)
\(b)\) \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)
\(\Leftrightarrow\)\(\left(2x-1\right)^{2010}.\left(2x-1\right)^2=\left(2x-1\right)^{2010}\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2=1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{2}\\x=\frac{0}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=1\)
Chúc bạn học tốt ~
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Tìm x:
a.\(\frac{1}{3}\times\left(x-1\right)+\frac{2}{5}\times\left(x+1\right)=0\)
b.\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}x-5\right)\)
c.\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
Các bn ơi giúp mk với chiều mk đi học rồi!!!!!!!!!!!!
Tim x biet :
\(a,\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(b,\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)
<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
<=> x = 2010
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\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)
Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)
=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)
=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay
\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16
=> -6+16 = 4x - 3x => x = 10
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