Tìm các số tự nhiên x sao cho :
1/ ( x + 2 ) ⋮ ( x + 1 )
2/ ( x + 5 ) ⋮ ( x + 1 )
3/ ( x + 4 ) ⋮ ( x + 3 )
4/ ( x + 3 ) ⋮ ( x + 2 )
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a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
a) Có: \(\left|x-2\right|\ge0\)
\(\left|x-10\right|\ge0\)
\(\Rightarrow\left|x-2\right|+\left|x-10\right|+4\ge4\)
Xét \(\orbr{\begin{cases}x-2=0\Rightarrow x=2\Rightarrow A=0+8+4=12\\x-10=0\Rightarrow x=10\Rightarrow A=8+0+4=12\end{cases}}\)
Vậy \(Min_A=12\) tại \(x=2\) hoặc \(10\)
b) Có: \(\left|x-1\right|\ge0\)
\(\left|x-2\right|\ge0\)
\(\left|x-3\right|\ge0\)
\(\Rightarrow B\ge0\)
Xét: \(\hept{\begin{cases}x-1=0\Rightarrow x=1\Rightarrow B=0+1+2=3\\x-2=0\Rightarrow x=2\Rightarrow B=1+0+1=2\\x-3=0\Rightarrow x=3\Rightarrow B=2+1+0=3\end{cases}}\)
Vậy \(Min_B=2\) tại \(x=2\)
b)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}:\frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(=\frac{1}{x-1}=\frac{1}{2009}\Leftrightarrow x+1=2009\)
\(\Rightarrow x=2009-1=2008\)
Bạn Phúc Trần Tấn bạn có biết làm phần a ko?Giúp mk với ạ!Mai mk cần rùi
Tìm x . biết :
\(a,\frac{2}{5}:\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
a. \(\frac{2}{5}.\left(-x-\frac{1}{2}\right)=\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}:\frac{4}{5}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{2}{5}.\frac{5}{4}\)
\(\Rightarrow-x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow-x=\frac{1}{2}+\frac{1}{2}\)
\(\Rightarrow-x=1\)
\(\Rightarrow x=-1\)
a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)
b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)
c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)
d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
2,
Ta có : \(\left(x+5\right)⋮\left(x+1\right)\)
\(\Leftrightarrow\frac{x+5}{x+1}\in N\Leftrightarrow\frac{x+1+4}{x+1}=\frac{x+1}{x+1}+\frac{4}{x+1}=1+\frac{4}{x+1}\)
Vì \(1\in N\)
\(\Leftrightarrow\frac{4}{x+1}\in N\Leftrightarrow x+1\inƯ_4=\left\{1;2;4\right\}\)
\(\Rightarrow x=\left\{0;1;3\right\}\)
mỏi tay quá ~ bạn làm nốt 2 ý còn lại nha .
1,
Ta có : \(\left(x+2\right)⋮\left(x+1\right)\)
\(\Leftrightarrow\frac{x+2}{x+1}\in N\Leftrightarrow\frac{x+1+1}{x+1}=\frac{x+1}{x+1}+\frac{1}{x+1}=1+\frac{1}{x+1}\)
Vì \(1\in N\)
\(\Rightarrow\frac{1}{n+1}\in N\Leftrightarrow n+1\inƯ_1=\left\{1\right\}\).
\(\Rightarrow n=\left\{0\right\}\)