\(\frac{x}{1,5}=\frac{3}{4,2}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
(1,5 + 3 + 4,5 + 6 + 7,5 + 9) x (4,2 x 10 : 8 + 4,2 x 8,75 - 4,2 : 0,1)
=31,5 x 0
=0
#Học Tốt
\(\left(1,5+3+4,5+6+7,5+9\right)\) x ( 4,2 x 10 : 8 + 4,2 x 8,75 - 4,2 : 0,1 )
\(=\left(1,5+3+4,5+6+7,5+9\right)\)x ( 4,2 x 1,25 + 4,2 x 8,75 - 4,2 : \(\frac{1}{10}\))
\(=\left(1,5+3+4,5+6+7,5+9\right)\)x ( 4,2 x 1,25 + 4,2 x 8,75 - 4,2 x 10 )
\(=\left(1,5+3+4,5+6+7,5+9\right)\)x 4,2 x ( 1,25 + 8,75 - 10 )
\(=\left(1.5+3+4.5+6+7,5+9\right)\)x 4,2 x 0 = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
(2,7+1,8)*2*1,5=4,5*2*1,5 =9*1,5 =13,5 (6/5*3/4)*2*2/3=9/10*2*2/3 =9/5*2/3 =6/5 2,7*1,8*1,5=7,29 6/5*3/4*2/3=3/5 16*16*6=1536
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2,7+1,8\right).2.1,5=13,5\)
\(\left(\frac{6}{5}.\frac{3}{4}\right).2.\frac{2}{3}=1,2\)
\(2,7.1,8.1,5=7,29\)
\(\frac{6}{5}.\frac{3}{4}.\frac{2}{3}=0,6\)
\(16.16.6=1536\)
= 13,5.
= \(\frac{6}{5}\).
= 7,29.
= \(\frac{3}{5}\).
= 1536.
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\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)
\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)
Vậy \(x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải :
Do \(VT\ge0\forall x;y\)nên ta có hệ :
\(\hept{\begin{cases}\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x=0\\1,5-\frac{11}{17}+\frac{23}{13}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{9}\\y=\frac{-377}{782}\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(ĐKXĐ:\hept{\begin{cases}x+y-3\ne0\\x-y+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne3-y\\x\ne y-1\end{cases}}}\)
Đặt: \(\hept{\begin{cases}x+y-3=a\\x-y+1=b\end{cases}}\)(1)
\(HPT\Leftrightarrow\hept{\begin{cases}\frac{5}{a}-\frac{2}{b}=8\\\frac{3}{a}+\frac{1}{b}=1,5\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{5}{a}-\frac{2}{b}=8\\\frac{6}{a}+\frac{2}{b}=3\end{cases}}\Leftrightarrow\frac{11}{a}=11\Leftrightarrow a=1}\)
Bn giải b xong rồi giải tiếp HPT (1)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 1,5 x 3 + ( 4,5 - 1,5 x 3 )
= 1,5.3 + ( 4,5 - 4,5)
= 1,5.3 + 0
= 4,5
b) \(\frac{15}{17}\).\(\frac{4}{33}\)+\(\frac{15}{17}\).\(\frac{29}{33}\)
\(=\)\(\frac{15}{17}\)\(\left(\frac{4}{33}+\frac{29}{33}\right)\)
=\(=\frac{15}{17}\)\(.1\)
\(=\frac{15}{17}\)
a) 1,5 x 3 + (4,5 - 1,5 x3)
=4,5 + 0
4,5
b) 15/17 x 4/33 + 15/17 x 29/33
= 15/17 x ( 4/33 + 29/33)
= 15/17 x 33/33
=15/17 x 1
= 1517
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
\(\frac{2x}{3}=\frac{5}{7}\Rightarrow2x.7=3.5\)
\(\frac{2x.7}{14}=\frac{3.5}{14}\)
\(x=\frac{3.5}{14}\)
\(\Rightarrow x=\frac{15}{14}\)