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a/ \(\left|-1,3\right|-\left|-3,7\right|+\left|-\dfrac{1}{2}\right|\)
\(=1,3-3,7+\dfrac{1}{2}\)
\(=-2,4+\dfrac{1}{2}\)
\(=-2,9\)
b/ \(\left|\dfrac{2}{5}\right|-\left|-0,2\right|.\left|-7\right|\)
\(=\dfrac{2}{5}-0,2.7\)
\(=\dfrac{2}{5}.1,4\)
\(=0,56\)
c/ \(-15:\left|-3\right|+\left|0,5\right|\)
\(=-15:2+0,5\)
\(=-7,5+0,5\)
\(=-8\)
Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?
Bạn kiểm tra lại nha
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)
<=> \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{18};x_2=-\dfrac{11}{18}\).
a/ \(\left|3x-1\right|=\left|5-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)
Vậy ......
b/ \(\left|x+2\right|-\left|x+7\right|=0\)
\(\Leftrightarrow\left|x+2\right|=\left|x+7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x+7\\x+2=-x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x=7-2\\x+x=-7-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\2x=-9\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{9}{2}\)
Vậy ...............
c/ \(\left|2x-1\right|+x=2\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2-x\\2x-1=-2+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+x=2+1\\2x-x=-2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy ..
Từ a/b=c/d⇒a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau
a/c=b/d=a+b/c+d
⇒a^3/c^3=b^3/d^3=(a+b)^3/(c+d)^3 (1)
Từ a^3/c^3=b^3/d^3=a^3-b^3/c^3-d^3 (2)
Từ (1) và (2)
⇒(a+b)^3/(c+d)^3=a^3-b^3/c^3-d^3
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
a/ \(\left|-x\right|=1,5\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)
Vậy .....
b/ \(\left|x+\dfrac{1}{2}\right|=2\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy ....
c/ \(\left|0,5-x\right|=\left|-0,5\right|\)
\(\left|0,5-x\right|=0,5\)
\(\Leftrightarrow\left[{}\begin{matrix}0,5-x=0,5\\0,5-x=-0,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy ...
Cảm ơn bn nha.