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\(a,x+13=5\\ \Rightarrow x=5-13\\ \Rightarrow x=-8\\ b,x-11=-18\\ \Rightarrow x=-18+11\\ \Rightarrow x=-7\)
a) Ta có: (x+1)(y-2)=-2
nên x+1; y-2 là các ước của -2
Trường hợp 1:
\(\left\{{}\begin{matrix}x+1=-1\\y-2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=4\end{matrix}\right.\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x+1=2\\y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x+1=-2\\y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=3\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x+1=1\\y-2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy: (x,y)\(\in\){(-2;4);(1;1);(-3;3);(0;0)}
b) Ta có: (x+1)(xy-1)=3
nên x+1;xy-1 là các ước của 3
Trường hợp 1:
\(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\Leftrightarrow loại\)
Trường hợp 2:
\(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Trường hợp 3:
\(\left\{{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)
Trường hợp 4:
\(\left\{{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-\dfrac{1}{2}\end{matrix}\right.\left(loại\right)\)
Vậy: \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;1\right)\right\}\)
c) Ta có: \(\left(x+y\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vây: (x,y)=(-1;1)
d) Ta có: \(\left|x+y\right|\cdot\left(x-y\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+y\right|=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=0\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy: (x,y)=(0;0)
a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\) b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
\(a,2x+34=56\\ \Rightarrow2x=56-34\\ \Rightarrow x=22:2\\ \Rightarrow x=11\\ b,87-\left(x-654\right):3=21\\ \Rightarrow\left(x-654\right):3=87-21\\ \Rightarrow x-654=66:3\\ \Rightarrow x=22+654\\ \Rightarrow x=676\\ c,7^{65}:7^x=7^{43}.7^{21}\\ \Rightarrow7^{65-x}=7^{43+21}\\ \Rightarrow65-x=64\\ \Rightarrow x=65-64\\ \Rightarrow x=1\)
b) Ta có: xy=-3
nên x,y là các ước của -3
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\left\{\left(1;-3\right);\left(-1;3\right);\left(-3;1\right);\left(3;-1\right)\right\}\)
\(a,2x+\dfrac{5}{2}=-\dfrac{3}{5}\)
\(2x=-\dfrac{3}{5}-\dfrac{5}{2}\)
\(2x=-\dfrac{31}{10}\)
\(x=-\dfrac{31}{10}:2\)
\(x=-\dfrac{31}{20}\)
\(b,\dfrac{1}{2}:x-\dfrac{5}{6}=-\dfrac{2}{3}\)
\(\dfrac{1}{2}:x=-\dfrac{2}{3}+\dfrac{5}{6}\)
\(\dfrac{1}{2}:x=\dfrac{1}{6}\)
\(x=\dfrac{1}{2}:\dfrac{1}{6}\)
\(x=3\)
\(c,\left(312-x\right):12,6=24,5\)
\(312-x=24,5\times12,6\)
\(312-x=308,7\)
\(x=312-308,7\)
`x=3,3`
a: =>2x=-3/5-5/2=-6/10-25/10=-31/10
=>x=-31/20
b: =>1/2:x=-2/3+5/6=5/6-4/6=1/6
=>x=1/2:1/6=3
c: =>312-x=308,7
=>x=3,3
a) \(8x+56:14=60\)
\(\Rightarrow8x+4=60\)
\(\Rightarrow8x=56\)
\(\Rightarrow x=\dfrac{56}{8}\)
\(\Rightarrow x=7\)
b) Mình làm rồi nhé !
c) \(41-2^{x+1}=9\)
\(\Rightarrow2^{x+1}=41-9\)
\(\Rightarrow2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
d) \(3^{2x-4}-x^0=8\)
\(\Rightarrow3^{2x-4}-1=8\)
\(\Rightarrow3^{2x-4}=9\)
\(\Rightarrow3^{2x-4}=3^2\)
\(\Rightarrow2x-4=2\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
g) \(65-4^{x+2}=2014^0\)
\(\Rightarrow65-4^{x+2}=1\)
\(\Rightarrow4^{x+2}=64\)
\(\Rightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=1\)
i) \(120+2\left(4x-17\right)=214\)
\(\Rightarrow2\left(4x-17\right)=214-120\)
\(\Rightarrow2\left(4x-17\right)=94\)
\(\Rightarrow4x-17=47\)
\(\Rightarrow4x=47+17\)
\(\Rightarrow4x=64\)
\(\Rightarrow x=16\)
a: \(8x+56:14=60\)
=>8x+4=60
=>8x=60-4=56
=>x=56/8=7
b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)
=>2x-3=3
=>2x=6
=>x=3
c: \(41-2^{x+1}=9\)
=>\(2^{x+1}=41-9=32\)
=>x+1=5
=>x=4
d: \(3^{2x-4}-x^0=8\)
=>\(3^{2x-4}-1=8\)
=>\(3^{2x-4}=8+1=9\)
=>2x-4=2
=>2x=6
=>x=3
g: \(65-4^{x+2}=2014^0\)
=>\(65-4^{x+2}=1\)
=>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
i: 120+2(4x-17)=214
=>2(4x-17)=214-120=94
=>4x-17=94/2=47
=>4x=64
=>\(x=\dfrac{64}{4}=16\)
`2x-15 = 17``
`=> 2x = 17 + 15`
`=> 2x = 32`
`=> X = 32 : 2`
`=> x = 16`
`156 - (x + 61) = 82`
`=> x + 61 = 156 - 82`
`=> x + 61 = 74`
`=> x = 13`
`2x - 138 = 2^3 . 3^2`
`=>2x - 138 = 72`
`=> 2x = 210`
`=> x = 105`
bài 2:
`23-3x = 8`
`=> 3x = 23 - 8`
`=> 3x = 15`
`=> x = 5`
`(x-35) - 120 = 0`
`=>(x-35) = 120`
`=> x = 120 +35`
`=> x = 155`
`3^x + 2 = 29`
`=> 3^x = 27`
`=> 3^x = 3^3`
`=> x = 3`