2x-5/9=1/3+[11/3-4+2/3] giúp mik với nha
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\(6\cdot x+9=15\)
\(\Rightarrow6\cdot x=15-9\)
\(\Rightarrow6\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{6}=1\)
_______________
\(43+2x=49\)
\(\Rightarrow2x=49-43\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
____________
\(x:2+5=11\)
\(\Rightarrow x:2=11-5\)
\(\Rightarrow x:2=6\)
\(\Rightarrow x=6\cdot2=12\)
______________
\(77-11x=0\)
\(\Rightarrow11x=77\)
\(\Rightarrow x=\dfrac{77}{11}\)
\(\Rightarrow x=7\)
_______________
\(12-4:x=8\)
\(\Rightarrow4:x=12-8\)
\(\Rightarrow4:x=4\)
\(\Rightarrow x=\dfrac{4}{4}=1\)
_____________
\(x:3+8=11\)
\(\Rightarrow x:3=11-8\)
\(\Rightarrow x:3=3\)
\(\Rightarrow x=3\cdot3=9\)
a: 6x+9=15
=>6x=6
=>x=1
b: 2x+43=49
=>2x=6
=>x=3
c: x:2+5=11
=>x:2=6
=>x=12
d: 77-11x=0
=>7-x=0
=>x=7
e: 12-4:x=8
=>4:x=4
=>x=1
f: x:3+8=11
=>x:3=3
=>x=9
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
mk ko viết đề nha:
a)=(1/2+1/2)+ (1/3+2/3)+(1/4 +3/4)+(1/5+4/5)
=(2/2 + 3/3) + (4/4 + 5/5)
=2/2 + 4/2
=6/4
À mik lm đc câu a thui nha!Nhưng k cho mik vs nhé!
c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)
\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)
\(=\dfrac{44}{75}\)
(\(\dfrac{2}{3}\) - \(\dfrac{11}{7}\) + \(\dfrac{13}{5}\)) - (5 - \(\dfrac{4}{7}\) + \(\dfrac{6}{5}\)) - (\(\dfrac{11}{3}\) - \(\dfrac{3}{5}\))
= \(\dfrac{2}{3}\) - \(\dfrac{11}{7}\) + \(\dfrac{13}{5}\) - 5 + \(\dfrac{4}{7}\) - \(\dfrac{6}{5}\) - \(\dfrac{11}{3}\) + \(\dfrac{3}{5}\)
= (\(\dfrac{2}{3}\) - \(\dfrac{11}{3}\)) - (\(\dfrac{11}{7}\) - \(\dfrac{4}{7}\)) + (\(\dfrac{13}{5}\) - \(\dfrac{6}{5}\)+ \(\dfrac{3}{5}\))
= \(\dfrac{-9}{3}\) - \(\dfrac{7}{7}\) + (\(\dfrac{7}{5}\) + \(\dfrac{3}{5}\)) - 5
= - 3 - 1 + 2 - 5
= - (3 + 1) - ( 5 - 2)
= - 4 - 3
= - 7
\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[\frac{11}{3}-4+\frac{2}{3}]\)
\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[-\frac{1}{3}+\frac{2}{3}]\)
\(2x-\frac{5}{9}=\frac{1}{3}+\frac{1}{3}\)
\(2x-\frac{5}{9}=\frac{2}{3}\)
\(2x=\frac{11}{9}\)
\(x=\frac{11}{18}\)
Vậy \(x=\frac{11}{18}\)