Tìm x, biết:
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
b) \(\frac{x-1}{x+5}=\frac{6}{7}\)
tìm x\(\in\) Q,biết:
a,\(\frac{x}{2}\left(\frac{3x}{5}\right)-\frac{13}{5}=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
b,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
c,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
d,\(\frac{13}{x-1}+\frac{5}{2x+2}-\frac{6}{3x-3}\)
e,\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Ai bt câu nào làm câu đấy nhé ! Làm hết cũng đc nhé!
a)\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
b)\(\left(\frac{-2}{3}x-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
c)\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
d)\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
e)\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}2\right)\)
k)\(\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
m)\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
n)\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Giải giúp mk
mk đang cần ghấp trong hôm nay
cám ơn mọi người
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
Bài 1. Giải các phương trình sau
1) \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}-2x\)
2) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
3) \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
4) \(\frac{2x+3}{3}=\frac{5-4x}{2}\)
5) \(\frac{5x+3}{12}=\frac{1+2x}{9}\)
6) \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
7) \(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
8) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2 \left(3x-1\right)}{5}-\frac{3x+2}{10}\)
9) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
10) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
tìm x biết:
a,\(7,5x:\left(9-6\frac{13}{21}\right)=2\frac{13}{25}\)
b,\(\frac{\left(1,16-x\right)\cdot5,25}{\left(10\frac{5}{9}-7\frac{1}{4}\right)\cdot2\frac{2}{17}}\)=75%
tìm x biết:
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}\)\(+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
B)\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{5}{2}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)
\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)
\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)
\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)
=> 5x - 20 = 2
=> 5x = 22
\(\Rightarrow x=\frac{22}{5}=4,4\)
Vậy, x = 4,4
a, \(\left(12-\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{13}=1\frac{2}{3}\)
b,\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right).2,8+1,75\right]:0,05=235\)
c.\(\frac{3-x}{5-x}=\left(\frac{3}{5}\right)^2\)
d,\(\left(1-\frac{3}{10}-x\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
e,\(\left(12\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{33}:\frac{8}{11}=1\frac{2}{3}\)
các bạn ơi giúp tớ với
\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
\(\left(\frac{-2}{3}x-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
Tìm x
\(\frac{5}{2}-\left(\frac{3}{2}-2\frac{1}{3}+x\right)=\frac{8}{15}-\left(\frac{1}{4}-\frac{7}{10}\right)\)
\(1\frac{2}{3}-1\frac{3}{5}+x=\frac{2}{5}-\left|\frac{3}{4}-\frac{7}{8}\right|\)
\(2-\left(\frac{2}{3}-3\frac{1}{4}+x\right)=1\left|\frac{1}{6}-\frac{13}{12}\right|\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
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