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nhân chéo là đc:
3(x+2)=-4(x-5)
3x+6=-4x+20
3x+4x=20-6
7x =14
x =2
Vậy x=2
= 10/3.x+ 67/4=-53/4
10/3.x=-53/4-67/4
10/3.x=-30
x=-30:10/3
x=-9
\(-52+\frac{2}{3}x=-46\)
\(\frac{2}{3}x=-46+52\)
\(\frac{2}{3}x=6\)
\(x=6:\frac{2}{3}\)
\(x=9\)
b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)
\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)
\(2,4.x-36=-\frac{12}{7}\)
\(2,4.x=-\frac{12}{7}+36\)
\(2,4.x=\frac{240}{7}\)
\(x=\frac{240}{7}\div2,4\)
\(x=\frac{100}{7}\)
\(\frac{1}{2}x-\frac{1}{4}=\frac{-1}{2}\)
\(\frac{1}{2}x=\frac{-1}{2}+\frac{1}{4}\)
\(\frac{1}{2}x=\frac{-2+1}{4}\)
\(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{-1}{4}:\frac{1}{2}\)
\(x=\frac{-1}{4}.2\)
\(x=\frac{-1}{2}\)
Vậy .....................
\(\frac{1}{2}x-\frac{1}{4}=-\frac{1}{2}\)
\(\frac{1}{2}x=-\frac{1}{2}+\frac{1}{4}\)
\(\frac{1}{2}x=-\frac{2}{4}+\frac{1}{4}\)
\(\frac{1}{2}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{1}{2}\)
\(x=-\frac{1}{4}\cdot2\)
\(x=-\frac{1}{2}\)
Vậy .....
\(\frac{3}{4}+\frac{1}{5}:\frac{7}{10}\)
\(=\frac{3}{4}+\frac{2}{7}\)
\(=\frac{29}{28}\)
\(\frac{3}{4}\)+ \(\frac{1}{5}\): \(\frac{7}{10}\)= \(\frac{3}{4}\)+ \(\frac{2}{7}\)
= \(\frac{28}{28}\)= 1
\(50\%x+\frac{2}{3}x=x-5\)
\(\Rightarrow\frac{1}{2}x+\frac{2}{3}x=x-5\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{2}{3}\right)=x-5\)
\(\Rightarrow x.\frac{7}{6}=x-5\)
\(\Rightarrow x-\frac{7}{6}x=5\)
\(\Rightarrow\frac{-x}{6}=5\Leftrightarrow-x=30\Leftrightarrow x=-30\)
\(50\%x+\frac{2}{3}x=x-5\)
\(\frac{1}{2}x+\frac{2}{3}x=x-5\)
\(x\left(\frac{1}{2}+\frac{2}{3}\right)=x-5\)
\(x.\frac{7}{6}=x-5\)
\(x.\frac{7}{6}-x=-5\)
\(x.\frac{1}{6}=-5\)
\(x=\left(-5\right):\frac{1}{6}=-30\)
Vậy x= -30
ta có: \(A=\frac{4n+1}{2n+3}=\frac{4n+6-5}{2n+3}=\frac{2.\left(2n+3\right)-5}{2n+3}=2-\frac{5}{2n+3}\)
Để A thuộc Z
=> 5/2n+3 thuộc Z
=> 5 chia hết cho 2n +3
=> 2n+3 thuộc Ư(5)={1;-1;5;-5}
nếu 2n + 3 = 1 => 2n = -2 => n = -1 (Loại)
2n+3 = -1 => 2n=-4 => n = -2 (Loại)
2n+3 = 5 => 2n = 2 => n = 1 (TM)
2n+3 = -5 => 2n = -8 => n = -4 (Loại)
\(\Rightarrow n\ne1\) thì A là phân số ( n thuộc N)
\(\frac{x+3}{-4}=-\frac{9}{x+3}\)
\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x+3\right)^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)
Vậy ....
quy đồng
\(\left(x+3\right)^2=36\)
\(\left(x+3\right)^2-6^2=0\)
áp dụng định lí " \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được
\(\left(x+3-6\right)\left(x+3+6\right)=0\)
\(x=3,x=-9\)