Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(0,\left(3\right)+3\dfrac{1}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{19}{45}\)
\(=\dfrac{184}{45}\)
b) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{1789}{990}\)
Bài 2:
a) \(0,\left(37\right)x=1\)
\(\Leftrightarrow\dfrac{37}{99}.x=1\)
\(\Leftrightarrow x=1:\dfrac{37}{99}\)
\(\Leftrightarrow x=\dfrac{99}{37}\)
b) \(0,\left(26\right)x=1,2\left(31\right)\)
\(\Leftrightarrow\dfrac{26}{99}x=\dfrac{1219}{990}\)
\(\Leftrightarrow x=\dfrac{1219}{990}:\dfrac{26}{99}\)
\(\Leftrightarrow x=\dfrac{1219}{260}\)
Chúc bạn học tốt!
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
Giải:
a) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{553}{330}-\dfrac{13}{99}\)
\(=\dfrac{139}{90}\)
Vậy giá trị của biểu thức trên là \(\dfrac{139}{90}\).
b) \(2\dfrac{1}{2}-3,4\left(12\right)-\dfrac{4}{3}\)
\(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}\)
\(=-\dfrac{301}{330}-\dfrac{4}{3}\)
\(=-\dfrac{247}{110}\)
Vậy giá trị của biểu thức trên là \(-\dfrac{247}{110}\).
Chúc bạn học tốt!
1.
a) \(-\dfrac{4}{9}+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}=-\dfrac{16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)
\(=-\dfrac{199}{36}\)
b) \(5\dfrac{1}{2}+\left(-3\right)=5\dfrac{1}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{5}{2}\)
c) \(4\dfrac{9}{11}+\left(-2\dfrac{1}{11}\right)=\dfrac{53}{11}-\dfrac{23}{11}=\dfrac{30}{11}\)
2.
a) \(4,3-\left(1,2\right)=3,1\)
b) \(0-\left(-0,4\right)=0+0,4=0,4\)
c) \(-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)
d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)
1.
a)\(-49+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}\)
\(=-49-\dfrac{5}{6}-\dfrac{17}{4}\)
\(=\dfrac{-588}{12}-\dfrac{10}{12}-\dfrac{51}{12}\)
\(=\dfrac{-588-10-51}{12}\)
\(=-\dfrac{649}{12}\)
b) \(5\dfrac{1}{2}+\left(-3\right)\)
\(=\dfrac{11}{2}-3\)
\(=\dfrac{11}{2}-\dfrac{6}{2}\)
\(=\dfrac{11-6}{2}\)
\(=\dfrac{5}{2}\)
c) \(4\dfrac{9}{11}+\left(2-2\dfrac{1}{11}\right)\)
\(=\dfrac{53}{11}+2-\dfrac{23}{11}\)
\(=\dfrac{53-23}{11}+2\)
\(=\dfrac{30}{11}+2\)
\(=\dfrac{30}{11}+\dfrac{22}{11}\)
\(=\dfrac{30+22}{11}\)
\(=\dfrac{52}{11}\)
2.
a) \(4,3-1,2=3,1\)
b) \(0-\left(-0,4\right)=0+0,4=0,4\)
c) \(-\dfrac{2}{3}-\dfrac{-1}{3}=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)
d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)
bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
a, \(\dfrac{13}{32}+\dfrac{8}{24}+\dfrac{19}{32}+\dfrac{2}{3}\)
\(=\left(\dfrac{13}{32}+\dfrac{19}{32}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{32}{32}+\dfrac{3}{3}=1+1=2\)
b, \(\dfrac{3}{4}.36\dfrac{1}{5}-\dfrac{3}{4}.2\dfrac{1}{5}\)
\(=\dfrac{3}{4}.\left(36\dfrac{1}{5}-2\dfrac{1}{5}\right)\)
\(=\dfrac{3}{4}.\left[\left(36-2\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\right]\)
\(=\dfrac{3}{4}.34=\dfrac{102}{4}=26\)
Bài 2:
a: x=27/10:9/5=27/10*5/9=135/90=3/2
b: =>|x|=1,75
=>x=1,75 hoặc x=-1,75
c: =>\(2-x=\sqrt[3]{25}\)
hay \(x=2-\sqrt[3]{25}\)
d: =>3^x-1*6=162
=>3^x-1=27
=>x-1=3
=>x=4
a: Ta có: \(0,\left(3\right)+\dfrac{10}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{4}{9}\)
\(=\dfrac{33}{9}+\dfrac{4}{9}=\dfrac{37}{9}\)
b: Ta có: \(\dfrac{4}{9}+1.2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{440}{990}+\dfrac{1219}{990}-\dfrac{130}{990}\)
\(=\dfrac{139}{90}\)
c: Ta có: \(2,\left(4\right)\cdot\dfrac{3}{11}\)
\(=\dfrac{22}{9}\cdot\dfrac{3}{11}\)
\(=\dfrac{2}{3}\)
d: Ta có: \(-0,\left(3\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
=0