\(\frac{x-3}{5}=\frac{5-2x}{-11}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,|x-\frac{1}{3}|+\frac{4}{5}=|-3,2+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=|-\frac{7}{10}|\)
\(\Rightarrow|x-\frac{1}{3}|+\frac{4}{5}=\frac{7}{10}\)
\(\Rightarrow|x-\frac{1}{3}|=\frac{7}{10}-\frac{4}{5}\)
\(\Rightarrow|x-\frac{1}{3}|=-\frac{1}{10}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{10}\\x-\frac{1}{3}=-\frac{1}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{30}\\x=\frac{7}{30}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đk: x \(\ne\)-2
Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)
<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10
<=> -4x - 8 = 5x + 10
<=> -4x - 5x = 10 + 8
<=> -9x = 18
<=> x = -2 (ktm)
=> pt vô nghiệm
b) Đk: x \(\ne\)2; x \(\ne\)-3
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
<=> x + 3 - 6x + 12 = -5
<=> -5x = -5 - 15
<=> -5x = -20
<=> x = 4
vậy S = {4}
c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11
Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)
<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)
Vậy S = {0}
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}\)
\(x=-\frac{1}{5}:5=-\frac{1}{25}\)
c) \(\frac{5}{3}-x=7+\frac{4}{5}\)
\(\frac{5}{3}-x=\frac{39}{5}\)
\(x=\frac{5}{3}-\frac{39}{5}=-\frac{92}{15}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}+\frac{5}{11}\)
\(2x=\frac{17}{22}\)
\(x=\frac{17}{22}:2\)
\(x=\frac{17}{44}\)
\(x=-\frac{1}{5}:5\)
NÈ BẠN!!!
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{25}{35}+\frac{28}{35}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}+\frac{11}{5}\)
\(5x=2\)
\(x=\frac{2}{5}\)
c)\(\frac{5}{3}-x=7\)
\(x=\frac{5}{3}-7=\frac{5}{3}-\frac{21}{3}=-\frac{16}{3}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}-\frac{-5}{11}=\frac{7}{22}-\frac{-10}{22}=\frac{17}{22}\)
\(x=\frac{17}{22}:2=\frac{17}{22}\cdot\frac{1}{2}=\frac{17}{44}\)
K CHO MÌNH NHA!!!
\(\Rightarrow\left(x-3\right).\left(-11\right)=\left(5-2x\right).5\)
\(\Rightarrow-11x+33=25-10x\)
\(\Rightarrow-11x-10x+33=25\)
\(\Rightarrow-21x=25-33\)
\(\Rightarrow-21x=-8\)
\(\Rightarrow x=\frac{8}{21}\)
\(\frac{x-3}{5}=\frac{5-2x}{-11}\)
\(\Leftrightarrow-11\left(x-3\right)=5\left(5-2x\right)\)
\(\Leftrightarrow33-11x=25-10x\)
\(\Leftrightarrow-11x+10x=25-33\)
\(\Leftrightarrow-x=-8\Leftrightarrow x=8\)