(4,5x - \(\frac{3}{4}\)* 5 \(\frac{1}{3}\)) * \(\frac{1}{12}\)+ \(\frac{1}{2}\)x = 1\(\frac{1}{2}\)
Tìm x giúp mình vớii
\(\left(4,5x-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=\frac{3}{2}\)
tìm x :
a)\(\left(4,5x-5\frac{1}{3}\right).\frac{1}{12}+50\%.x=\frac{3}{2}\)
b)60%x + 2/3 x=684
giải giùm mk nha các bn
a (\(\frac{9}{2}\)x - \(\frac{16}{3}\)).\(\frac{1}{12}\) + \(\frac{1}{2}\)x=\(\frac{3}{2}\)
\(\frac{9}{2}\)x . \(\frac{1}{12}\) - \(\frac{16}{3}\) . \(\frac{1}{12}\) + \(\frac{1}{2}\)x=\(\frac{3}{2}\)
( \(\frac{9}{2}\)x . \(\frac{1}{12}\) + \(\frac{1}{2}\)x) - \(\frac{16}{3}\) . \(\frac{1}{12}\) = \(\frac{3}{2}\)
x. ( \(\frac{9}{2}\) . \(\frac{1}{12}\) +\(\frac{1}{2}\)) - \(\frac{4}{9}\) = \(\frac{3}{2}\)
x.\(\frac{7}{8}\) = \(\frac{3}{2}\) + \(\frac{4}{9}\) = \(\frac{35}{18}\)
x= \(\frac{35}{18}\) : \(\frac{7}{8}\) = \(\frac{20}{9}\) Vậy x=\(\frac{20}{9}\)
b 60%+2/3x=684
3/5x+2/3x=684
x(3/5+ 2/3) = 684
x. 19/15 = 684
x=540. Vậy x=540
a) \(-\frac{2}{5}+\frac{2}{3}.x+\frac{1}{6}.x=-\frac{4}{5}\)
b)\(\frac{3}{2}.x-\frac{2}{5}-\frac{2}{3}.x=-\frac{4}{15}\)
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
Giải giúp mình với, được câu nào thì làm câu nấy ạ! Mình cần gấp
a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25
b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
Cảm ơn bạnh nha. Chúc bạn buổi tối ấm =)))) <3
Thực hiện phép tính
P = \(\frac{3}{x^4-x^3+x-1}\)- \(\frac{1}{x^4+x^3-x-1}\)- \(\frac{4}{x^5-x^4+x^3-x^2+x-1}\)
GIÚP MÌNH VỚII
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
1 Tìm x
a/-1\(\frac{4}{12}\)-1\(\frac{3}{12}\)<_x<_\(\frac{4}{3}.\frac{13}{7}\)
b/(\(\frac{x}{-5}+\frac{2}{3}\)):2\(\frac{1}{5}-1,4.\frac{15}{49}=\frac{-5}{21}\)
c/\(\frac{-2}{3}+\frac{-5}{12}< x< _-4-\frac{1}{3}:\frac{1}{6}\)
d/(\(\frac{2}{3}x-\frac{1}{2}\)).\(\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)
2 Viết số thời gian sau dưới dạng hỗn số có đơn vị giờ
a/2h45ph
b/1h45ph
c/85ph
mai mình nộp nên mong mọi người giúp mình
1,Tính
a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}\)
b,\(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
2,Tìm x
a,\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
Giúp mình với nhé mai mình phải nộp rồi
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!
1.
a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=5+1+\frac{1}{2}=\frac{13}{2}\)
2.
a) \(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
\(\frac{2}{3}x=-\frac{1}{4}\)
\(x=-\frac{3}{8}\)
b) \(\frac{3}{5}+\frac{2}{5}:x=3\frac{1}{2}\)
\(\frac{2}{5}x=\frac{7}{2}-\frac{3}{5}\)
\(\frac{2}{5}x=\frac{29}{10}\)
\(x=\frac{29}{4}\)
c) \(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
\(\left(\frac{x+4}{2004}+1\right)+\left(\frac{x+3}{2005}+1\right)=\left(\frac{x+2}{2006}+1\right)\left(\frac{x+1}{2007}+1\right)\)
\(\frac{x+2008}{2004}+\frac{x+2008}{2005}-\frac{x+2008}{2006}-\frac{x+2008}{2007}=0\)
\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)
\(x+2008=0\)
\(x=-2008\) ( vì \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\))
\(\frac{4}{9}:\frac{-1}{7}+6\frac{5}{9}.\frac{2}{3}\)
\(\left(\frac{-1}{3}\right)^4.\frac{4}{11}+\frac{7}{11}.\left(\frac{-1}{3}\right)^2\)
\(\left(\frac{-1}{7}\right)^0-2\frac{4}{9}:\left(\frac{2}{3}\right)^2\)
\(\left(\frac{1}{3}-\frac{5}{6}\right)^2+\frac{5}{6}:2\)
Tìm x
\(\left(\frac{3}{5}-\frac{1}{2}x\right)^2+8=12\)
\(\frac{7}{12}-\left|\frac{1}{3}-\frac{1}{2}x\right|=\frac{1}{3}\)
\(\frac{13}{12}-\frac{1}{3}.\left|\frac{1}{5}-\frac{1}{3}x\right|=\frac{1}{2}\)
Nhanh hộ mình nha
Cảm ơn mn