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Bài 1:
a) Ta có: \( \left|x+2\right|=\left|3-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=3-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-3-2\\x+2x=3-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-5\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left|2x-4\right|=\left|3-x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=3-x\\2x-4=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+4\\2x-x=-3+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=7\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=1\end{matrix}\right.\)
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Phân tích :
\(10=2.5\)
\(12=2^2.3\)
\(15=3.5\)
\(BCNN\left(10;12;15\right)=2^23.5=60\)
Ta có:10 = 2.5
12 = 22. 3
15 = 3.5
BCNN (10,12,15) = 22 .3.5 = 60
Vậy x = 60
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1: x.(x+7) = 0
Th1:x=0 Th2:x+7=0
=>x=-7
bài 2 (x+12).(x-3)= 0
Th1:x+12=0 Th2:x-3=0
=>x=-12 =>x=3
bài 3 (-x+5).(3-x)=0
Th1 (-x)+5=0 Th2:3-x=0
=>-x=-5 =>x=3
bài 4 x.(2+x).(7-x)=0
Th1:x=0 Th3:7-x=0
Th2:2+x=0 =>x=7
=>x=-2
bài 5 (x-1).(x+2).(-x-3)=0
Th1:x-1=0 Th2:x+2=0
=>x=1 =>x=-2
Th3:-x-3=0
=>-x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{121}{35}.\left(\frac{1}{7}-\frac{5}{21}\right).\left|-\frac{11}{7}\right|\)
\(=\frac{121}{35}.\left(\frac{3}{21}-\frac{5}{21}\right).\frac{11}{7}\)
\(=\frac{121}{35}.\frac{-2}{21}.\frac{11}{7}\)
\(=-\frac{242}{735}.\frac{11}{7}\)
\(=-\frac{2662}{5145}\)
Chúc bạn học tốt !!!
Xem lại đề nhé !!! Số to quá !!!
\(\frac{121}{35}\times\left(\frac{1}{7}-\frac{5}{21}\right)\times\left|\frac{-11}{7}\right|\)
\(=\frac{121}{35}\times\frac{-2}{21}\times\frac{11}{7}\)
\(=\frac{-2662}{5145}\)
Chúc bn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)2\cdot3\cdot4+4\cdot5\cdot6\\ =4\cdot\left(2\cdot3+5\cdot6\right)⋮4\)
\(2)3\cdot5\cdot7-1\cdot3\cdot5\\ =5\cdot\left(3\cdot7-1\cdot5\right)⋮5\)
\(3)4\cdot6\cdot8-2\cdot3\cdot4\\ =4\cdot\left(6\cdot8-2\cdot3\right)⋮4\)
\(4)4\cdot6\cdot8+11\cdot12\cdot13\\ =\left(4\cdot6\right)\cdot8+11\cdot12\cdot13\\ =24\cdot8+11\cdot12\cdot13\\ =12\cdot2\cdot8+11\cdot12\cdot13\\ =12\cdot\left(2\cdot8+11\cdot13\right)⋮12\)
\(5)5\cdot6\cdot7+33\cdot34\cdot35\\ =\left(5\cdot7\right)\cdot6+33\cdot34\cdot35\\ =35\cdot6+33\cdot34\cdot35\\ =35\cdot\left(6+33\cdot34\right)⋮35\)
\(6)43\cdot45\cdot47-9\cdot7\cdot5\\ =43\cdot45\cdot47-\left(9\cdot5\right)\cdot7\\ =43\cdot45\cdot47-45\cdot7\\ =45\cdot\left(43\cdot47-7\right)⋮45\)
\(7)125\cdot13+50\cdot14\\ =25\cdot5\cdot13+25\cdot2\cdot14\\ =25\cdot\left(5\cdot13+2\cdot14\right)⋮25\)
\(8)125\cdot13+50\cdot14\)
Có: 125*13 ⋮ 13
Mà: \(50\cdot14=5^2\cdot2\cdot2\cdot7=5^2\cdot2^2\cdot7\) => không chia hết cho 13
=> 125*13 + 50*14 không chia hết cho 13