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25 tháng 7 2017
1.
a) \(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b) \(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
c) \(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}=\frac{23}{2}-\frac{65}{2}=-21\)
d) \(\left(-49,1\right).\frac{13}{27}-58,9.\frac{13}{27}=\frac{13}{27}.\left(-49,1-58,9\right)=\frac{13}{27}.\left(-108\right)=-52\)
e) \(0,375:\left(-4,5\right)=\frac{-1}{12}\)
f) \(3\frac{1}{7}:\left(-1\frac{3}{7}\right)=\frac{22}{7}:\frac{-10}{7}=\frac{-11}{5}\)
g) \(9\frac{1}{3}:4\frac{2}{3}-2=\frac{28}{3}:\frac{14}{3}-2=2-2=0\)
h) \(\left(7\frac{3}{4}:0,3125+4,5.2\frac{2}{45}\right):\left(-8,5\right)=\left(\frac{31}{4}:\frac{5}{16}+\frac{9}{2}.\frac{92}{45}\right):\frac{-17}{2}=\left(\frac{124}{5}+\frac{46}{5}\right):\frac{-17}{2}=34:\frac{-17}{2}=-4\)
25 tháng 7 2017
Bài 1 : Tính:
a)
\(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b)
\(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{10}{15}+\frac{-2}{15}=\frac{8}{15}\)
c)
\(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}\)\(=\frac{23}{2}-\frac{65}{2}=\frac{-42}{2}=-21\)
....
Tự lm tiếp dạng như v
Bài 2 :
\(A=\frac{-6}{11}.\frac{7}{10}.\frac{11}{-6}.-20=\left(\frac{-6}{11}.\frac{11}{-6}\right).\left(\frac{7}{10}.-20\right)\)\(=1.\left(-14\right)=-14\)
.....
Bài 3 :
\(\frac{3}{7}.x-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{3}{7}-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{1}{35}x=\frac{-17}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}:\frac{1}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}.35=-17\)
TD
2
16 tháng 6 2015
A = 6 -2/3 + 1 /2 - 5 -5/3 +3/2 -3 + 7/3 - 5/2
= (6 - 5 - 3) - ( 2/3 -5/3 + 7/3 ) + ( 1/2 +3/2 - 5/2)
= -2 + 0 -1/2 = -5/2
24 tháng 7 2017
\(\left|x-y-2\right|+\left|y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|x-y-2\right|\ge0\forall x;y\\\left|y+3\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-2\right|+\left|y+3\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-y-2\right|=0\Rightarrow x-\left(-3\right)-2=0\Rightarrow x+1=0\Rightarrow x=-1\\\left|y+3\right|=0\Rightarrow y+3=0\Rightarrow y=-3\end{matrix}\right.\)
\(\left|x-2007\right|+\left|y-2008\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2007\right|\ge0\forall x\\\left|y-2008\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-2007\right|+\left|y-2008\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2007\right|=0\Rightarrow x-2007=0\Rightarrow x=2007\\\left|y-2008\right|=0\Rightarrow y-2008=0\Rightarrow y=2008\end{matrix}\right.\)
\(\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|=0\)
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|\ge0\forall x\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|=0\Rightarrow\dfrac{1}{6}+\dfrac{3}{4}x=0\Rightarrow\dfrac{3}{4}x=-\dfrac{1}{6}\Rightarrow x=-\dfrac{2}{9}\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|=0\Rightarrow\dfrac{29}{34}+\dfrac{23}{13}x=0\Rightarrow\dfrac{23}{13}x=-\dfrac{29}{34}\Rightarrow x=-\dfrac{377}{782}\end{matrix}\right.\)
\(\left|x-y-5\right|+\left|y-2\right|\le0\)
\(\left\{{}\begin{matrix}\left|x-y-5\right|\ge0\forall x;y\\\left|y-2\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|+\left|y-2\right|\le0\\\left|x-y-5\right|+\left|y-2\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\Rightarrow x-2-5=0\Rightarrow x=7\\\left|y-2=0\right|\Rightarrow y=2\end{matrix}\right.\)
\(\left|3x+2y\right|+\left|4y-1\right|\le0\)
\(\left\{{}\begin{matrix}\left|3x+2y\right|\ge0\forall x;y\\ \left|4y-1\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|3x+2y\right|+\left|4y-1\right|\ge0\\\left|3x+2y\right|+\left|4y-1\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x+2y\right|=0\Rightarrow3x+\dfrac{1}{2}=0\Rightarrow3x=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{6}\\\left|4y-1\right|=0\Rightarrow4y=1\Rightarrow y=\dfrac{1}{4}\end{matrix}\right.\)
16 tháng 6 2019
\(\left|\frac{2}{3}-\frac{1}{2}+\frac{3}{4x}\right|+\left|1,5-\frac{11}{17}+\frac{23}{13y}\right|=0\)
Phân tích :
ta biết khi hai số đối nhau cộng lại sẽ có tổng bằng 0 , nhưng trong trường hợp trên thì sẽ không có 2 số đối nhau bởi vì cả hai phép tích đều có giá trị tuyệt đối
=> \(\left|\frac{2}{3}-\frac{1}{2}+\frac{3}{4x}\right|=\left|1,5-\frac{11}{17}+\frac{23}{13y}\right|=0\)
=>\(\frac{2}{3}-\frac{1}{2}+\frac{3}{4x}=\frac{3}{2}-\frac{11}{17}+\frac{23}{13y}=0\)
=> \(\frac{1}{6}+\frac{3}{4x}=\frac{29}{34}+\frac{23}{13y}=0\)
=> \(\left[{}\begin{matrix}\frac{1}{6}+\frac{3}{4x}=0\\\frac{29}{34}+\frac{23}{13y}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\frac{3}{4x}=0-\frac{1}{6}=-\frac{1}{6}\\\frac{23}{13y}=0-\frac{29}{34}=-\frac{29}{34}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}4x=3.6:\left(-1\right)=-18\\13y=23.34:\left(-29\right)=-\frac{782}{29}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-18:4=-\frac{9}{2}\\y=-\frac{782}{29}:13=-\frac{782}{377}\end{matrix}\right.\)
Vậy (x;y)∈{\(-\frac{9}{2};-\frac{782}{377}\)}
a) 1,(3) + 0,(23)
=4/3+23/99
=155/99
b) 12,0(3) - 11,(2)
=361/30-101/9
=73/90