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Ta có: \(\frac{7}{10}-\frac{5}{8}.x-\frac{7}{6}=\frac{5}{6}.x\)
\(\Rightarrow\frac{7}{10}-\frac{7}{6}=\frac{5}{6}x+\frac{5}{8}x\)
\(\Rightarrow\frac{21}{30}-\frac{35}{30}=\left(\frac{5}{6}+\frac{5}{8}\right)x\)
\(\Rightarrow-\frac{7}{15}=\left(\frac{15}{24}+\frac{20}{24}\right)x\)
\(\Rightarrow-\frac{7}{15}=\frac{35}{24}x\)
\(\Rightarrow x=-\frac{7}{15}:\frac{35}{24}\)
\(\Rightarrow x=-\frac{7}{15}.\frac{24}{35}\)
\(\Rightarrow x=-\frac{8}{25}\)
Vậy \(x=-\frac{8}{25}\)
Chuk bạn hok tốt! 
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)
\(\Leftrightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}\right)\)
\(\Leftrightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)
ta chuyển về vế trái được
\(\Leftrightarrow\left(x+18\right)\left(\frac{1}{13}+\frac{1}{122}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
\(\Leftrightarrow x+2018=0\)(do cái còn lại khác 0)
\(\Leftrightarrow x=-2018\)
mình nghĩ đề cậu viết thiếu mình sửa rồi
Ta có:
\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)
\(\Rightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}+1\right)\)
\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)
\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}-\frac{x+18}{10}-\frac{x+18}{9}-\frac{x+18}{8}=0\)
\(\Rightarrow\left(x+18\right)\times\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Vì \(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
\(\Rightarrow x+18=0\)
\(\Rightarrow x=-18\)
Vậy phương trình có nghiệm là x = -18
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
2:
\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)
B=(x1+x2)^2-2x1x2
=3^2-2*(-7)
=9+14=23
C=căn (x1+x2)^2-4x1x2
=căn 3^2-4*(-7)=căn 9+28=căn 27
D=(x1^2+x2^2)^2-2(x1x2)^2
=23^2-2*(-7)^2
=23^2-2*49=431
D=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=69+10*(-7)=-1
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
B nha
Chọn B nhé
Lần sau e lưu ý đăng bên môn toán để được trợ giúp nhanh hơn nhé