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a. 1440 : [ 120 - (3x + 9 ) ] = 120
120 - (3x + 9) = 1440 : 120 = 12
3x + 9 = 120 - 12 = 108
3x = 108 - 9 = 99
x = 99 : 3 = 33
b. 120 + [ ( 999 - 9x ) : 60 ] . 24 = 480
[ ( 999 - 9x ) : 60 ] . 24 = 480 - 120 = 360
( 999 - 9x ) : 60 = 360 : 24 = 15
( 999 - 9x ) = 15 . 60 = 900
9x = 999 - 900 = 99
x = 99 : 9 = 11
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Câu 1:
=>5x=15
hay x=3
Câu 2:
\(\Leftrightarrow120-\left(3x+9\right)=12\)
=>3x+9=108
=>3x=99
hay x=33
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\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Leftrightarrow\left(4-x\right)\left(4-x\right)=-5\times-5\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Rightarrow\left(4-x\right).\left(4-x\right)=\left(-5\right).\left(-5\right)\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left(4-x\right)^2=5^2\)
\(\Rightarrow4-x=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4-5\\x=4-\left(-5\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
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Bài 1:
a, 96 \(⋮x=>x\inư\left(96\right)\)
b, \(2^x.15+2^x.17=4^{30}\)
\(2^x\left(15+17\right)=4^{30}\)
\(2^x.32=4^{30}\)
\(2^x.2^5=2^{60}\)
\(2^x=2^{60}:2^5\)
\(2^x=2^{55}\)
=> x = 55
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\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
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\(\frac{1440}{120-\left(3x+9\right)}=120\)
\(\Leftrightarrow120-\left(3x+9\right)=\frac{1440}{120}=12\)
\(\Leftrightarrow3x+9=120-12\)
\(\Leftrightarrow3x+9=108\)
\(\Leftrightarrow3x=108-9=99\)
\(\Leftrightarrow x=99\div3=33\)
vậy x=33
1440 / [ 120 - ( 3x + 9 ) ] =120
120 - ( 3x + 9 ) = 1440 : 120
120 - ( 3x + 9 ) = 12
3x + 9 = 120 - 12
3x + 9 = 108
3x = 108 - 9
3x = 99
x = 99 : 3
x = 33
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì: \(ƯCLN\left(x,y\right)=9\)
\(\Rightarrow\hept{\begin{cases}x=9.k_1\\y=9.k_2\end{cases}}\)
Mà: \(ƯCLN\left(k_1,k_2\right)=1\)
\(\Rightarrow9.k_1.9.k_2=1215\)
\(\Rightarrow k_1.k_2=15\)
Ta có bảng sau:
\(k_1\) | \(1\) | \(3\) |
\(k_2\) | \(15\) | \(5\) |
Nếu: \(k_1=1\Rightarrow k_2=15\Rightarrow\hept{\begin{cases}x=9\\y=135\end{cases}}\)
Nếu: \(k_1=3\Rightarrow k_2=5\Rightarrow\hept{\begin{cases}x=27\\y=45\end{cases}}\)
Vậy: \(\left(x,y\right)=\left(9;135\right),\left(27;45\right)\)
4(x-12)-2=50
4(x-12)=50+2
4(x-12)=52
x-12=52÷4
x-12=13
x=25
4(x-12)-2=50
4(x-12)=50+2
4(x-12)=52
x-12=52÷4
x-12=13
x=25