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chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
a) \(\dfrac{3}{5}\) . x = \(\dfrac{21}{10}\)
x = \(\dfrac{21}{10}\) : \(\dfrac{3}{5}\)
x = \(\dfrac{21}{10}\) . \(\dfrac{5}{3}\)
x = \(\dfrac{105}{30}\)
x = \(\dfrac{7}{2}\)
b) \(\dfrac{x}{20}\) = \(\dfrac{4}{5}\)
\(\dfrac{x}{20}\) = \(\dfrac{16}{20}\)
=> x = 16
Bạn sai câu b) rồi. Câu b) ta có công thức: a:b=c:d => a.d=b.c mới đúng vì đây là tỉ lệ thức chứ sao bạn áp dụng giống như quy đồng thế?
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy...
\(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a. \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=-1+4=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
Vậy..........
b. \(1\dfrac{3}{4}.x+1\dfrac{1}{2}=-\dfrac{4}{5}\)
\(\Rightarrow1\dfrac{3}{4}x=-\dfrac{4}{5}-1\dfrac{1}{2}=\dfrac{-23}{10}\)
\(\Rightarrow x=\dfrac{-23}{10}:1\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{-46}{35}\)
a) \(4^{x+2}.3^x=16.12^5\)
\(\Leftrightarrow4^{x+2}.3^x=4^2.4^5.3^5\)
\(\Leftrightarrow4^{x-5}.3^{x-5}=1\)
\(\Leftrightarrow12^{x-5}=1\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
\(a)4^{x+2}.3^x=4^2.\left(4.3\right)^5\)
\(4^{x+2}.3^x=4^7.3^5\)
\(\Rightarrow4^{x+2}=4^7;3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
1: Để \(\dfrac{-5}{x-1}< 0\) thì x-1>0
hay x>1
2: Để \(\dfrac{7}{x-6}>0\) thì x-6>0
hay x>6
3: Để \(\dfrac{-3}{x-6}< 0\) thì x-6<0
hay x<6
Câu a)
\(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Leftrightarrow2x-2-6x-6-8x-12=16\)
\(\Leftrightarrow-12x-20=16\)
\(\Leftrightarrow-12x=16+20\)
\(\Leftrightarrow-12x=36\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3\)
Câu c)
\(\dfrac{2x-y}{5}=\dfrac{3x-2z}{15}\) \(\Rightarrow\dfrac{3\left(2x-y\right)}{3\times5}=\dfrac{3x-2z}{15}\) \(\Rightarrow\dfrac{6x-3y}{15}=\dfrac{3x-2z}{15}\)
\(\Rightarrow6x-3y=3x-2z\)
\(\Rightarrow6x-3x+2z=3y\)
\(\Rightarrow3x+2z=3y\)
\(\Rightarrow\left(2x+2z\right)+x=3y\)
\(\Rightarrow2\left(x+z\right)+x=3y\)
\(\Rightarrow2\times2y+x=3y\)
\(\Rightarrow4y+x=3y\)
\(\Rightarrow x=3y-4y\)
\(\Rightarrow x=-y\)