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a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\\ x^8=x^7\\ \Rightarrow x=1;x=-1\)
b)\(x^{10}=25.x^8\\ x^2=25\\ \Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)
\(\Rightarrow x^8=x^7\)
\(\Rightarrow x^8-x^7=0\)
\(\Rightarrow x^7.x-x^7=0\)
\(\Rightarrow x^7\left(x-1\right)=0\)
\(\Rightarrow x-1=0\) (vì x^7 \(\ne\)0)
\(\Rightarrow\) x=1
b) x^10=25x^8
\(\Rightarrow x^8.x^2-25x^8=0\)
\(\Rightarrow x^8\left(x^2-25\right)=0\)
\(\Rightarrow x^8=0\) hoặc \(x^2-25=0\)
1) x^8=0
\(\Rightarrow\) x=0(1)
2) x^2 -25=0
x^2=0+25
x^2=25
x^2=5^2 hay x^2=(-5)^2
Suy ra x=5 hoặc x=-5 (2)
Từ (1) và (2)\(\Rightarrow\)x\(\in\left\{0;5;-5\right\}\)
EM KO CHÉP ĐÁP ÁN NHÉ
a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
\(\Rightarrow x^8=x^7\)
\(\Rightarrow x^8-x^7=0\)
\(\Rightarrow x^7.\left(x-1\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(x^7\ne0\) )
Vậy \(x=1\)
b ) \(x^{10}=25x^8\)
\(\Rightarrow x^{10}-25x^8=0\)
\(\Rightarrow x^8.\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8=0\) hoặc \(x^2-25=0\)
Do đó \(x=0\) hoặc \(x=5\) hoặc \(x=-5\)
Vậy \(x\in\left\{0;5;-5\right\}\)
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(\Leftrightarrow\dfrac{2}{x-3}-\dfrac{2}{x-2}+\dfrac{1}{x-8}-\dfrac{1}{x-3}+\dfrac{1}{x-20}-\dfrac{1}{x-8}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{2}{x-2}=\dfrac{-3}{4}\)
\(\Leftrightarrow4\left(x-2\right)-8\left(x-3\right)=-3\left(x-3\right)\left(x-2\right)\)
\(\Leftrightarrow4x-8-8x+24+3\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow3x^2-15x+18-4x+16=0\)
\(\Leftrightarrow3x^2-19x+34=0\)
\(\text{Δ}=\left(-19\right)^2-4\cdot3\cdot34=-47< 0\)
Do đó: Phương trình vô nghiệm
\(\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)
=> 3x -1 = 2/3
3x = 5/3
x = 5/9
học tốt ^^
\(\left(x^4\right)^2=x^{12-5}\)
\(x^8-x^7=0\)
\(x^7\cdot x-x^7=0\)
\(x^7\cdot\left(x-1\right)=0\)
+) x^7 = 0 => x = 0
+) x -1 = 0 => x = 1
Vậy,...........
học tốt ^^
Giải:
a) Ta có:
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\left(x^7\ne0\right)\Leftrightarrow x=1\)
Vậy \(x=1\)
b) Ta có:
\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow\) \(\left[\begin{array}{}x^8=0\\x^2-25=0\end{array}\right.\)
\(\Leftrightarrow\) \(\left[\begin{array}{}x=0\\x=5\\x=-5\end{array}\right.\) Vậy...
cảm ơn bn