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a, => x^2+5 = 0
=> x^2=-5 ( vô lí vì x^2 >= 0)
=> ko tồn tại x tm bài toán
b, Vì x^2-5 > x^2-25
Mà (x^2-5): (x^2-25) < 0
=> x^2-5 >0 và x^2-25 <0
=> 5 < x^2 < 25
=> \(x>\sqrt{5}\)hoặc \(x< -\sqrt{5}\) và -5 < x < 5
=> -5 < x < -\(\sqrt{5}\)hoặc \(\sqrt{5}\)< x < 5
k mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x2+5).(x2-25)=0
trường hợp 1(+):
x2+5=0
x2=0-5
x2=-5
không có giá trị nguyên nào
trường hợp 2
x2-25=0
x2=0+25
x2=25
x=5
vậy x =5
câu b làm tương tự nha. thông cảm
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a: =>5(x+2)=170
=>x+2=34
=>x=32
b: =>x+2=5 hoặc x+2=-5
=>x=-7 hoặc x=3
c: =>x-2=1 hoặc x-2=-1
=>x=1 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$x(x+\frac{2}{5})=\frac{8}{25}$
$x^2+\frac{2}{5}x-\frac{8}{25}=0$
$25x^2+10x-8=0$
$(5x+4)(5x-2)=0$
$\Rightarrow 5x+4=0$ hoặc $5x-2=0$
$\Rightarrow x=\frac{-4}{5}$ hoặc $x=\frac{2}{5}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
b, \(\left[{}\begin{matrix}5-x=-5\\5-x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)
a. \(\left(x-1\right)^2=1\)
\(\Leftrightarrow\left(x-1\right)^2=\left(\pm1\right)^2\)
TH1: \(x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\)
TH2: \(x-1=\left(-1\right)\)
\(\Leftrightarrow x=\left(-1\right)+1\)
\(\Leftrightarrow x=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
Vậy \(x=2\) hoặc \(x=0\)
b. \(\left(5-x\right)^2=25\)
\(\Leftrightarrow\left(5-x\right)^2=\left(\pm5\right)^2\)
TH1: \(5-x=5\)
\(\Leftrightarrow x=5-5\)
\(\Leftrightarrow x=0\)
TH2: \(5-x=\left(-5\right)\)
\(\Leftrightarrow x=5-\left(-5\right)\)
\(\Leftrightarrow x=10\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}\)
Vậy \(x=0\) hoặc \(x=10\)
HT
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5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 25+x=0
x=0-25
x=-25
b)\(2^x:2^{19}=2^{25}\)
\(2^x=2^{25}.2^{19}\)
\(2^x=2^{44}\)
=>x=44
c)\(5^x.5^{18}=5^{54}\)
\(5^x=5^{54}:5^{18}\)
\(5^x=5^{36}\)
=>x=36
![](https://rs.olm.vn/images/avt/0.png?1311)
\(25\%x+x=\frac{5}{2}\)
\(\frac{1}{4}x+1x=\frac{5}{2}\)
\(\left(\frac{1}{4}+1\right)x=\frac{5}{2}\)
\(\frac{5}{4}x=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{5}{4}\)
\(x=2\)
Thank you bạn nhé