Toán Bài 1:tìm x a)(2x-3)²=9 b)(x-1)mũ 7=(x-1)³ c)7²-(13+4x(=5.2³ 14/08/2021 By Melanie Bài 1:tìm x a)(2x-3)²=9 b)(x-1)mũ 7=(x-1)³ c)7²-(13+4x(=5.2³
a)$(2x-3)²=9^{}$ =>$(2x-3)²=(-3)²=3²^{}$ =>\(\left[ \begin{array}{l}2x-3=-3\\2x-3=3\end{array} \right.\) =>\(\left[ \begin{array}{l}2x=0\\2x=6\end{array} \right.\)=> \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) b) $(x-1)^{7}$ =$(x-1)^{3}$ => $TH1$: \(\left[ \begin{array}{l}(x-1)^{7}=0\\(x-1)^{3}=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) =>$TH2$: \(\left[ \begin{array}{l}(x-1)^{7}=1\\(x-1)^{3}=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x=2\\x=2\end{array} \right.\) =>$TH3$: \(\left[ \begin{array}{l}(x-1)^{7}=-1\\(x-1)^{3}=-1\end{array} \right.\) =>\(\left[ \begin{array}{l}x=0\\x=0\end{array} \right.\) c)$7^{2}$ -$(13+4x)=5.2^{3}$ =>$49-(13+4x)=40$ =>$(13+4x)=49-40$ =>$(13+4x)=9$ =>$4x=-4$ =>$x=-1$ Trả lời
a) $(2x-3)^2=9$ \(⇒\left[ \begin{array}{l}2x-3=3\\2x-3=-3\end{array} \right.\) \(⇒\left[ \begin{array}{l}2x=6\\2x=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=3\\x=0\end{array} \right.\) b) $(x-1)^7=(x-1)^3$ \(⇒\left[ \begin{array}{l}(x-1)^7=0⇒x=1\\(x-1)^3=0⇒x=1\end{array}⇒x=1 ™ \right.\) \(⇒\left[ \begin{array}{l}(x-1)^7=1⇒x=2\\(x-1)^3=1⇒x=2\end{array}⇒x=2(tm) \right.\) \(⇒\left[ \begin{array}{l}(x-1)^7=-1⇒x=0\\(x-1)^3=-1⇒x=0\end{array}⇒x=0(tm) \right.\) c) $7^2-(13+4x)=5.2^3$ $49-(13+4x)=5.8=40$ $13+4x=49-40=9$ $4x=9-13=-4$ $⇒x=-1$ d) $x^5=x^3$ \(⇒\left[ \begin{array}{l}x^5=1⇒x=1\\x^3=1⇒x=1\end{array}⇒x=1 \right.\) \(⇒\left[ \begin{array}{l}x^5=-1⇒x=-1\\x^3=-1⇒x=-1\end{array}⇒x=-1 \right.\) \(⇒\left[ \begin{array}{l}x^5=0⇒x=0\\x^3=0⇒x=0\end{array}⇒x=0 \right.\) Trả lời
a)$(2x-3)²=9^{}$
=>$(2x-3)²=(-3)²=3²^{}$
=>\(\left[ \begin{array}{l}2x-3=-3\\2x-3=3\end{array} \right.\)
=>\(\left[ \begin{array}{l}2x=0\\2x=6\end{array} \right.\)=> \(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
b) $(x-1)^{7}$ =$(x-1)^{3}$
=> $TH1$: \(\left[ \begin{array}{l}(x-1)^{7}=0\\(x-1)^{3}=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
=>$TH2$: \(\left[ \begin{array}{l}(x-1)^{7}=1\\(x-1)^{3}=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x=2\\x=2\end{array} \right.\)
=>$TH3$: \(\left[ \begin{array}{l}(x-1)^{7}=-1\\(x-1)^{3}=-1\end{array} \right.\) =>\(\left[ \begin{array}{l}x=0\\x=0\end{array} \right.\)
c)$7^{2}$ -$(13+4x)=5.2^{3}$
=>$49-(13+4x)=40$
=>$(13+4x)=49-40$
=>$(13+4x)=9$
=>$4x=-4$
=>$x=-1$
a) $(2x-3)^2=9$
\(⇒\left[ \begin{array}{l}2x-3=3\\2x-3=-3\end{array} \right.\)
\(⇒\left[ \begin{array}{l}2x=6\\2x=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=3\\x=0\end{array} \right.\)
b) $(x-1)^7=(x-1)^3$
\(⇒\left[ \begin{array}{l}(x-1)^7=0⇒x=1\\(x-1)^3=0⇒x=1\end{array}⇒x=1 ™ \right.\)
\(⇒\left[ \begin{array}{l}(x-1)^7=1⇒x=2\\(x-1)^3=1⇒x=2\end{array}⇒x=2(tm) \right.\)
\(⇒\left[ \begin{array}{l}(x-1)^7=-1⇒x=0\\(x-1)^3=-1⇒x=0\end{array}⇒x=0(tm) \right.\)
c) $7^2-(13+4x)=5.2^3$
$49-(13+4x)=5.8=40$
$13+4x=49-40=9$
$4x=9-13=-4$
$⇒x=-1$
d) $x^5=x^3$
\(⇒\left[ \begin{array}{l}x^5=1⇒x=1\\x^3=1⇒x=1\end{array}⇒x=1 \right.\)
\(⇒\left[ \begin{array}{l}x^5=-1⇒x=-1\\x^3=-1⇒x=-1\end{array}⇒x=-1 \right.\)
\(⇒\left[ \begin{array}{l}x^5=0⇒x=0\\x^3=0⇒x=0\end{array}⇒x=0 \right.\)