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a/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ..............
b, \(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{13}{39}< \dfrac{13}{38}\)
\(\Leftrightarrow\dfrac{13}{38}>\dfrac{-12}{-37}\)
a)\(\text{|}x+\dfrac{3}{4}\text{|}-\dfrac{1}{3}=0\)
=>\(\text{|}x+\dfrac{3}{4}\text{|}=\dfrac{1}{3}\)
=>\(x+\dfrac{3}{4}=-\dfrac{1}{3}\)hoặc\(x+\dfrac{3}{4}=\dfrac{1}{3}\)
=>\(x=-\dfrac{13}{12}\)hoặc\(x=-\dfrac{5}{12}\)
Vậy...
b)\(\dfrac{13}{38}\) và \(\dfrac{-12}{-37}\)
Ta có:\(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{1}{3}=\dfrac{13}{39}< \dfrac{13}{38}\)
=>\(\dfrac{13}{38}>\dfrac{-12}{-37}\)
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT 
\(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=39/13+5/13=44/13
1.Tìm x, biết:
x/3=y/5
Theo tính chất dãy tỉ số bằng nhau ta có:
x+y/3+5= 16/8=2
=>x=6; y=10
2.Cho a+5/a−5=b+6/b−6(a≠5;b≠6)
CMR: ab=56
Giải:
ta có a+5/a-5=b+6/b-6 =>a+5/b+6=a-5/b-6 (*)
=> a+5+a-5/b+6+b-6=2a/2b=a/b (1)
Lại có: (*)=a+5-a+5/b+6-b+6=10/12=5/6 (2)
Từ 1 và 2 suy ra a/b=5/6 (đpcm)
\(P=\sqrt{\left(x-\dfrac{3}{4}\right)^2}+\dfrac{1}{4}\)
\(=\left|x-\dfrac{3}{4}\right|+\dfrac{1}{4}\)
Ta có : \(\left|x-\dfrac{3}{4}\right|\ge0\forall x\Rightarrow\left|x-\dfrac{3}{4}\right|+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
\(\Rightarrow P\ge\dfrac{1}{4}\)
Dấu "=" xảy ra
\(\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
Vậy GTNN của P là \(\dfrac{1}{4}\) khi x = \(\dfrac{3}{4}\)
\(\dfrac{\dfrac{2}{3}+0,25-0,6}{\dfrac{2}{3}-0,25+0,6}:\dfrac{\dfrac{2}{5}-\dfrac{1}{6}+\dfrac{3}{7}}{\dfrac{2}{5}+\dfrac{1}{6}-\dfrac{3}{7}}\)
\(=\dfrac{\dfrac{19}{60}}{\dfrac{61}{60}}:\dfrac{\dfrac{139}{210}}{\dfrac{29}{210}}=\dfrac{19}{61}:\dfrac{139}{29}=\dfrac{551}{8479}\)
Mình không chắc lắm!! Chúc bạn học tốt!!!
1: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=5/13+39/13
=44/13
2: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)
A= \(\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}\)+ \(\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{9}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
A= \(\dfrac{5}{13}\)+ \(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}{1\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}\)
A= \(\dfrac{5}{13}+3\) = \(\dfrac{44}{13}\)