Tìm x biết : \(2006\times\left|x-1\right|+\left(x-1\right)^2=2005\times\left|1-x\right|\)
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Tìm x biết : \(\left(x+1\right)\times\left(x^2+x+1\right)\times\left(x-1\right)\times\left(x^2-x+1\right)=7\)
(x+1) * (x2 +x+1) * (x-1) * (x2-x+1) = 7
[(x+1) * (x2 +x+1) ]*[(x-1) * (x2-x+1)]= 7 [Áp dụng hằng đẳng thức a3+b3=(a+b)*(a2+ab+b2)]
(x3+13) * (x3-13) = 7
x3 * x3 - x3 * 13 + x3 * 13 - 13 *13 =7
(x3)2 - 1 = 7
(x3)2 =7+1
(x3)2 =8
suy ra x = 3 căn 2
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Bài 2: Tìm x:a)dfrac{x-1}{27}dfrac{-3}{1-x} c)3times x2times y vàx-2times y8b)dfrac{4}{5}-left|x-dfrac{1}{2}right|dfrac{3}{4} d)dfrac{x-1}{2005}dfrac{3-y}{2006} và x-4009y
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Bài 2: Tìm x:
a)\(\dfrac{x-1}{27}\)=\(\dfrac{-3}{1-x}\) c)\(3\times x=2\times y\) và\(x-2\times y=8\)
b)\(\dfrac{4}{5}\)-\(\left|x-\dfrac{1}{2}\right|\)=\(\dfrac{3}{4}\) d)\(\dfrac{x-1}{2005}\)=\(\dfrac{3-y}{2006}\) và x-4009=y
a: \(\Leftrightarrow\left(x-1\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
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Tìm x \(\frac{2}{\left(x-1\right)\times\left(x-3\right)}+\frac{5}{\left(x-3\right)\times\left(x-8\right)}+\frac{12}{\left(x-8\right)\times\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)-3/4
Tìm x biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
cho: \(f\left(x\right)=a\times x^3+4\times x\times\left(x^2-1\right)+8\)
\(g\left(x\right)=x^3-4\times x\times\left(b\times x+1\right)+c-3\)
tìm a, b, c để \(f\left(x\right)=g\left(x\right)\)
Tìm x biết
a)\(x^4-30x^2+31x-30=0\)
b)\(\left(x^2+x\right)^2+4\times\left(x^2+x\right)=12\)
c)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
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Tìm x thuộc Z biết
a. left(x-1right)timesleft(2x-4right)0
b. left(x^2+5right)timesleft(x-5right)0
c. left(x^2+5right)timesleft(x^2-2right)0
d. left(x^2-1right)timesleft(x^2-9right) 0
e. left|x^2-2xright|x
g. left|2x+1right|+left|x+8right|4x
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Tìm x thuộc Z biết
a. \(\left(x-1\right)\times\left(2x-4\right)=0\)
b. \(\left(x^2+5\right)\times\left(x-5\right)=0\)
c. \(\left(x^2+5\right)\times\left(x^2-2\right)=0\)
d. \(\left(x^2-1\right)\times\left(x^2-9\right)< 0\)
e. \(\left|x^2-2x\right|=x\)
g. \(\left|2x+1\right|+\left|x+8\right|=4x\)
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
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Tìm x:
\(\left(1-\frac{1}{2}\right)\)\(\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right).200x=4036\)
\(\frac{1}{2}.\frac{2}{3}....\frac{99}{100}.200x=4036\)
\(\frac{1.2.....99}{2.3...100}.200x=4036\)
\(\frac{1}{100}.20x=4036\)
\(\frac{20x}{100}=4036\)
\(x=2018\)
Dấu ''.'' là dấu nhân nhé
Xem thêm câu trả lời
Tìm x biết 2006 . \(\left|x-1\right|\)+\(^{\left(x-1\right)^2}\) = 2005 . \(\left|1-x\right|\)
Tập hợp giá trị x thõa mãn là
\(2006.\left|x-1\right|+\left(x-1\right)^2=2005.\left|1-x\right|\) (để thỏa mản là chúng bằng nhau thì ta cần tích của chúng bằng 0)
Ta tính vế phải:
\(2005.\left|x-1\right|=0\)
\(\left|x-1\right|=0\)
Ta có: x - 1 = 0
=> x = 0 + 1 = 1
Mà vế trái bằng vế phải nên x = 1
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