Bài 1:tìm x a)(2x-3)²=9 b)(x-1)mũ 7=(x-1)³ c)7²-(13+4x(=5.2³

Question

Bài 1:tìm x
a)(2x-3)²=9
b)(x-1)mũ 7=(x-1)³
c)7²-(13+4x(=5.2³

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Melanie 4 tháng 2021-08-14T19:57:00+00:00 2 Answers 6 views 0

Answers ( )

    0
    2021-08-14T19:58:19+00:00

    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)$7^{2}$ -$(13+4x)=5.2^{3}$ 

    =>$49-(13+4x)=40$ 

    =>$(13+4x)=49-40$ 

    =>$(13+4x)=9$ 

    =>$4x=-4$ 

    =>$x=-1$ 

    0
    2021-08-14T19:58:21+00:00

    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) $x^5=x^3$

    \(⇒\left[ \begin{array}{l}x^5=1⇒x=1\\x^3=1⇒x=1\end{array}⇒x=1 \right.\)

    \(⇒\left[ \begin{array}{l}x^5=-1⇒x=-1\\x^3=-1⇒x=-1\end{array}⇒x=-1 \right.\)

    \(⇒\left[ \begin{array}{l}x^5=0⇒x=0\\x^3=0⇒x=0\end{array}⇒x=0 \right.\)

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