Tìm x, biết:
a) \(\frac{1+2+...+50}{2+4+...+50}-x=\frac{250}{150}+\frac{15}{9}\)
b) 7+9+11+...+2011+2.x=6+8+10+...+100
c) \(\frac{8}{5}-x=3.x-\frac{7}{3}\)
d) 6.x+2.x+x=100
(Phiền các bạn ghi ra cách giải giùm mình)
K
Khách
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Những câu hỏi liên quan
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
13 tháng 9 2017
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
HT
13 tháng 9 2017
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
18 tháng 3 2020
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
18 tháng 3 2020
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
7 tháng 3 2016
Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9
\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)
b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)
<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)
<=>x=-43/30
Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại
7 tháng 3 2016
a,A=1863/623:2/7
A=1863/178
b,
ta có:
1863/178.x+5/6=-3/4
1863/178.x=-19/12
=>x=-1691/11178
28 tháng 3 2020
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
3 tháng 4 2016
(x+2)/17+(x+4)/15+(x+6)/13=(x+8)/11+(x+10)/9+(x+12)/7
=>(x+2+17)/17+(x+4+15)/15+(x+6+13)/13=(x+8+11)/11+(x+10+9)/9+(x+12+7)/7
=>(x+19)/17+(x+19)/15+(x+19)/13=(x+19)/11+(x+19)/9+(x+19)/7
=>(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0
=>(x+19)*(1/17+1/15+1/13-1/11-1/9-1/7)=0
=>x+19=0
=>x=19
TN
3 tháng 4 2016
áp dụng tc tỉ lệ thức ta có :
\(\Leftrightarrow\frac{671x+2804}{3315}=\frac{239x+2462}{693}\Rightarrow\left(671x+2804\right)693=3315\left(239x+2462\right)\)
=>(671x+2804)693=693(671x+2804) (VT)
<=>693(671x+2804)=3315(239x+2462)
=>465003x+1943172=792285x+8161530
=>-327282x=621835
=>x=621835:(-327282)
=>x=-19
16 tháng 2 2021
a) \(x+\left(-7\right)=-20\)
\(\Rightarrow x=-20+7\)
\(\Rightarrow x=-13\)
Vậy \(x=-13\)
b) \(8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
c) \(|x|-7=-6\)
\(\Rightarrow|x|=-6+7\)
\(\Rightarrow|x|=1\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy \(x\in\left\{1;-1\right\}\)
d) \(5^2.2^2-7.|x|=65\)
\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)
\(\Rightarrow10^2-7.|x|=65\)
\(\Rightarrow100-7.|x|=65\)
\(\Rightarrow7.|x|=35\)
\(\Rightarrow|x|=5\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy \(x\in\left\{5;-5\right\}\)
e) \(37-3.|x|=2^3-4\)
\(\Rightarrow37-3.|x|=8-4\)
\(\Rightarrow37-3.|x|=4\)
\(\Rightarrow3.|x|=33\)
\(\Rightarrow|x|=11\)
\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)
Vậy \(x\in\left\{11;-11\right\}\)
f) \(|x|+|-5|=|-37|\)
\(\Rightarrow|x|+5=37\)
\(\Rightarrow|x|=32\)
\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)
Vậy \(x\in\left\{32;-32\right\}\)
g)\(5.|x+9|=40\)
\(\Rightarrow|x+9|=8\)
\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)
Vậy \(x\in\left\{-1;-17\right\}\)
h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
Vậy \(-3\le x\le4\)
