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Giải:
Ta có:
\(\left(a+b+c+d\right)-\left(a+c+d\right)._{\left(1\right)}\)
\(=a+b+c+d-a-c-d.\)
\(=\left(a-a\right)+\left(c-c\right)+\left(d-d\right)+b.\)
\(=0+0+0+b=b.\)
Thay số vào \(_{\left(1\right)}\)\(\Rightarrow1-2=b\Rightarrow b=-1\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+d\right)._{\left(2\right)}\)
\(=a+b+c+d-a-b-d.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(d+d\right)+c.\)
\(=0+0+0+c=c.\)
Thay số vào \(_{\left(2\right)}\)\(\Rightarrow1-3=c\Rightarrow c=-2\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+c\right)_{\left(3\right)}.\)
\(=a+b+c+d-a-b-c.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(c-c\right)+d.\)
\(=0+0+0+d=d.\)
Thay số vào \(_{\left(3\right)}\)\(\Rightarrow1-4=d\Rightarrow d=-3\in Z.\)
\(\Rightarrow a+b+c+d=1.\)
\(a+\left(-1\right)+\left(-2\right)+\left(-3\right)=1.\)
\(\Rightarrow a=1-\left(-1\right)-\left(-2\right)-\left(-3\right).\)
\(\Rightarrow a=1+1+2+3=7\in Z.\)
Vậy \(\left\{a;b;c;d\right\}=\left\{7;-1;-2;-3\right\}.\)
Do a + b + c + d = 1 mà a + c + d = 2
=> b = 1 - 2 = -1
=> c = 1 - 3 = -2
=> d = 1 - 4 = -3
=> a = 1 - (-1 - 2 - 3) = 7
@Valentine
\(A=\frac{11}{9}-\frac{7}{8}+-\frac{2}{3}-\frac{1}{8}+\frac{25}{9}-\frac{4}{3}\)
\(A=1\)
\(B=1\frac{3}{4}:\frac{3}{5}-\frac{2}{3}x1,75+\left(\frac{1}{2}\right)^2:\frac{1}{7}\)
\(B=3,5\)
a) \(\frac{a}{b}x-\frac{7}{8}=\frac{1}{4}\)
\(\Rightarrow\frac{a}{b}x=\frac{1}{4}+\frac{7}{8}\)
\(\Rightarrow\frac{a}{b}x=\frac{9}{8}\)
\(\Rightarrow x=\frac{9}{8}:\frac{a}{b}=\frac{9}{8}.\frac{b}{a}\)
\(\Rightarrow x=\frac{9b}{8a}\)
b) \(\frac{3}{2}x-\frac{1}{2}=\frac{1}{3}:\left(\frac{-5}{6}\right)\)
\(\Rightarrow\frac{3}{2}x-\frac{1}{2}=\frac{-2}{5}\)
\(\Rightarrow\frac{3}{2}x=\frac{-2}{5}+\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\frac{3}{2}\)
\(\Rightarrow x=\frac{1}{15}\)
c) \(\frac{2}{3}\left(x+\frac{5}{4}\right)-\frac{1}{3}\left(\frac{2}{3}-x\right)=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{5}{6}-\frac{2}{9}+\frac{1}{3}x=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{1}{3}x=\frac{4}{3}-\frac{5}{6}+\frac{2}{9}\)
\(\Rightarrow x=\frac{13}{18}\)
Bài 1:
a) \(-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-0,125\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-5}{2}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\left(\frac{-20}{8}-\frac{1}{8}\right)\\ =-\frac{4}{5}-\frac{8}{25}\cdot\frac{-21}{8}\\ =-\frac{4}{5}-\frac{-21}{25}\\ =\frac{-4}{5}+\frac{21}{25}\\ =\frac{-20}{25}+\frac{21}{25}=\frac{1}{25}\)
c) \(5\frac{1}{2}-4\frac{2}{3}:\frac{16}{9}-3\frac{1}{3}:\frac{16}{9}\\ =5\frac{1}{2}-\left(4\frac{2}{3}:\frac{16}{9}+3\frac{1}{3}:\frac{16}{9}\right)\\ =5\frac{1}{2}-\left(4\frac{2}{3}+3\frac{1}{3}\right):\frac{16}{9}\\ =5\frac{1}{2}-8\cdot\frac{9}{16}\\ =\frac{11}{2}-\frac{9}{2}=\frac{2}{2}=1\)
Bài 2:
a) \(\left(20\%x+\frac{2}{5}x-2\right):\frac{1}{3}=-2013\\ \left(\frac{1}{5}x+\frac{2}{5}x-2\right)\cdot3=-2013\\ \left[x\left(\frac{1}{5}+\frac{2}{5}\right)-2\right]=\left(-2013\right):3\\ x\cdot\frac{3}{5}-2=-671\\ x\cdot\frac{3}{5}=-671+2\\ x\cdot\frac{3}{5}=-669\\ x=\left(-669\right):\frac{3}{5}\\ x=\left(-669\right)\cdot\frac{5}{3}\\ x=-1115\)Vậy x = -1115
b) \(\left(4,5-2\left|x\right|\right)\cdot1\frac{4}{7}=\frac{11}{14}\\ \left(\frac{9}{2}-2\left|x\right|\right)\cdot\frac{11}{7}=\frac{11}{14}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}:\frac{11}{7}\\ \frac{9}{2}-2\left|x\right|=\frac{11}{14}\cdot\frac{7}{11}\\ \frac{9}{2}-2\left|x\right|=\frac{1}{2}\\ 2\left|x\right|=\frac{9}{2}-\frac{1}{2}\\ 2\left|x\right|=4\\ \left|x\right|=4:2\\ \left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)Vậy x ∈ {2 ; -2}
Ta có: \(\frac{2}{3}a=\frac{1}{4}b\)
\(\Leftrightarrow\frac{2a}{3}=\frac{b}{4}\)
\(\Leftrightarrow2a=\frac{3b}{4}\)
hay \(a=\frac{3b}{4}:2=\frac{3b}{8}\)
Ta có: \(\frac{1}{2}b=\frac{1}{3}c\)
\(\Leftrightarrow\frac{b}{2}=\frac{c}{3}\)
hay \(c=\frac{3b}{2}\)
Ta có: a+b+c=90
\(\Leftrightarrow\frac{3b}{8}+b+\frac{3b}{2}=90\)
\(\Leftrightarrow b\left(\frac{3}{8}+1+\frac{3}{2}\right)=90\)
\(\Leftrightarrow b\cdot\frac{23}{8}=90\)
hay \(b=90:\frac{23}{8}=\frac{720}{23}\)
Ta có: \(a=\frac{3b}{8}\)(cmt)
hay \(a=3\cdot\frac{720}{23}:8=\frac{270}{23}\)
Ta có: a+b+c=90
\(\Leftrightarrow c=90-a-b=90-\frac{270}{23}-\frac{720}{23}=\frac{1080}{23}\)
Vậy: \(\left(a,b,c\right)=\left(\frac{270}{23};\frac{720}{23};\frac{1080}{23}\right)\)