1) Viết các số sau dưới dạng lũy thừa

a) \(625=5^4\)

b) \(\frac{4}{9}=\left(\frac{2}{3}\right)^2\)

c) \(0,81=0,9^2\)

d) \(\frac{9}{64}=\left(\frac{3}{8}\right)^2\)

\(625=5^4\)

\(\frac{4}{9}=\left(\frac{2}{3}\right)^2\)

\(0,81=\left(0,9\right)^2\)

\(\frac{9}{64}=\left(\frac{3}{8}\right)^2\)

a.\(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

c,\(\frac{25}{49}=\left(\frac{5}{7}\right)^2\)

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

2.

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

⇒ \(5x+1=\pm\frac{6}{7}\)

⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

a, \(\left(x+1\right)^2=169\)

\(\left(x+1\right)^2=13^2\)

\(x+1=13\)

\(x=13-1\)

\(x=12\)

1.

a) \(\left(x+1\right)^2=169\)

⇒ \(x+1=\pm13\)

⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)

Vậy \(x\in\left\{12;-14\right\}.\)

b) \(\left(x+3\right)^3=-\frac{1}{27}\)

⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)

⇒ \(x+3=-\frac{1}{3}\)

⇒ \(x=\left(-\frac{1}{3}\right)-3\)

⇒ \(x=-\frac{10}{3}\)

Vậy \(x=-\frac{10}{3}.\)

c) \(\left(2x-4\right)^4=\frac{1}{625}\)

⇒ \(2x-4=\pm\frac{1}{5}\)

⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)

Còn câu d) bạn làm tương tự như mấy câu trên.

Chúc bạn học tốt!

Bài 1:

a) Ta có: \(MN^2+MP^2=8^2+15^2=289\)

Mà \(NP^2=17^2=289\)

Nên \(MN^2+MP^2=NP^2\) \(\Rightarrow\Delta MNP\) vuông tại \(M.\)(đpcm)

b) Xét \(\Delta MNI\) và \(\Delta KNI\) có:

\(\widehat{NMI}=\widehat{NKI}=90^0\)

\(NI:\) cạnh chung

\(\widehat{MNI}=\widehat{KNI}\left(g.t\right)\)

\(\Rightarrow\Delta MNI=\Delta KNI\left(đpcm\right)\)

c) Ta có: \(\widehat{NIM}=\widehat{NIK}\left(\Delta MNI=\Delta KNI\right)\)

\(\widehat{MIQ}=\widehat{KIP}\) (đối đỉnh)

\(\Rightarrow\widehat{NIQ}=\widehat{NIP}\left(1\right)\)

Xét \(\Delta NIQ\) và \(\Delta NIP\) có:

\(\widehat{QNI}=\widehat{PNI}\left(g.t\right)\)

\(NI:\) cạnh chung

\(\widehat{NIQ}=\widehat{NIP}\left(1\right)\)

\(\Rightarrow\Delta NIQ=\Delta NIP\left(g.c.g\right)\)

\(\Rightarrow IQ=IP\left(2\right)\)

Xét \(\Delta MIQ\) và \(\Delta KIP\) có:

\(\widehat{IMQ}=\widehat{IKP}=90^0\)

\(\widehat{NIQ}=\widehat{NIP}\left(1\right)\)

\(IQ=IP\left(2\right)\)

\(\Rightarrow\Delta MIQ=\Delta KIP\) (cạnh huyền - góc nhọn)

\(\Rightarrow MQ=KP\left(đpcm\right)\)

Bài 3:

a) \(\text{Áp dụng định lí Pi-ta-go vào }\Delta\text{ ABC vuông tại A, ta có:}\)

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=3^2+4^2\)

\(\Leftrightarrow BC^2=25\)

\(\Rightarrow BC=\sqrt{25}=5\left(cm\right)\)

b) \(\Delta ABD\text{ là tam giác vuông cân, vì:}\)

\(\widehat{BAD}=90^0\)

\(AB=AD\)

c) \(\text{Ta có: }\)\(\left\{{}\begin{matrix}AD=AB\\AC=AE\end{matrix}\right.\) \(\Rightarrow AD+AC=AB+AE\Rightarrow DC=BE\left(1\right)\)

\(\text{Xét }\Delta\text{ ACE có: }\)

\(AC=AE\)

\(\Rightarrow\Delta ACE\text{ cân tại A}\)

\(\Rightarrow\widehat{ACE}=\widehat{AEC}\left(2\right)\)

\(\text{Xét }\Delta CDE\text{ và }\Delta EBC\text{ có:}\)

\(DC=BE\left(1\right)\)

\(\widehat{ACE}=\widehat{AEC}\left(2\right)\)

\(EC\text{: cạnh chung}\)

\(\Rightarrow\Delta CDE=\Delta EBC\left(c.g.c\right)\)

\(\Rightarrow DE=BC\left(đpcm\right).\)

!