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\(\left(x-5\right)^5=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)
\(\Rightarrow\left(x-5\right)^5.\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^5.\left(x-6\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^5=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
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** Bổ sung điều kiện $x,y$ là các số nguyên.
$x+5y+xy=6$
$(x+xy)+5y=6$
$x(1+y)+5(y+1)=11$
$(y+1)(x+5)=11$
Vì $x,y$ nguyên nên $x+5, y+1$ cũng nguyên. Ta xét các TH sau:
TH1: $x+5=1, y+1=11\Rightarrow x=-4; y=10$
TH2: $x+5=11, y+1=1\Rightarrow x=6; y=0$
TH3: $x+5=-1; y+1=-11\Rightarrow x=-6; y=-12$
TH4: $x+5=-11; y+1=-1\Rightarrow x=-16; y=-2$
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Bài 4:
a: =>7/x-5=2
=>x-5=7/2
=>x=17/2
b: =>1-2x=-5
=>2x=6
=>x=3
c: =>2x-3=5 hoặc 2x-3=-5
=>2x=8 hoặc 2x=-2
=>x=-1 hoặc x=4
d: =>2(x+1)^2+17=21
=>2(x+1)^2=4
=>(x+1)^2=2
=>\(x+1=\pm\sqrt{2}\)
=>\(x=\pm\sqrt{2}-1\)
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\(a.\left[x\cdot\left(x+1\right)\right]:2=136\\ x\cdot\left(x+1\right)=136\cdot2\\ x\cdot\left(x+1\right)=2\cdot2\cdot2\cdot17\cdot2\\ x\cdot\left(x+1\right)=\left(2\cdot2\cdot2\cdot2\right)\cdot17\\ x\cdot\left(x+1\right)=16\cdot17\\ =>x=16\\ b.\left[x\cdot\left(x+1\right)\right]:2=300\\ x\cdot\left(x+1\right)=3\cdot2\cdot2\cdot5\cdot5\cdot2\\ x\cdot\left(x+1\right)=\left(3\cdot2\cdot2\cdot2\right)\cdot\left(5\cdot5\right)\\ x\cdot\left(x+1\right)=24\cdot25\\ =>x=24\\ c.\left[x\cdot\left(x+1\right)\right]:2=561\\ x\cdot\left(x+1\right)=2\cdot3\cdot11\cdot17\\ x\cdot\left(x+1\right)=\left(17\cdot2\right)\cdot\left(3\cdot11\right)\\ x\cdot\left(x+1\right)=34\cdot33\\ =>x=33\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{2}=\frac{x}{4}\)
\(\Leftrightarrow\)\(\frac{2\left(x-1\right)}{4}=\frac{x}{4}\)
\(\Leftrightarrow\)\(2\left(x-1\right)=x\)
\(\Leftrightarrow\)\(2x-2=x\)
\(\Leftrightarrow\)\(2x-x=2\)
\(\Leftrightarrow\)\(x=2\)
Vậy....
\(\frac{x+1}{x-1}=\frac{1}{2}\)
\(\Leftrightarrow\)\(2\left(x+1\right)=x-1\)
\(\Leftrightarrow\)\(2x+2=x-1\)
\(\Leftrightarrow\)\(2x-x=-1-2\)
\(\Leftrightarrow\)\(x=-3\)
Vậy....
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a, (x-1)x+2 = (x-1)x+6
<=> (x-1)x . (x-1)2 = (x-1)x . (x-1)6
<=> (x-1)2 = (x-1)6
<=> \(\orbr{\begin{cases}x-1=0\\x-1=1\end{cases}}\)
<=>\(\orbr{\begin{cases}x=0+1\\x=1+1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy x thuộc {1,2}
b, 19.x2+x = 0
<=> x2+x = 0
<=> x2 . xx = 0
<=> x = 0
Vậy x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3^{x-1}+3x+3^{x+1}=1053\)
\(=3^x:3+3^x+3^x.3=1053\)
\(=3^x.\dfrac{1}{3}+1+3=1053\)
\(=3^x.\dfrac{13}{5}=1053\)
\(=3^x=243\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
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