hình như bạn chụp thiếu đúng k

chụp lại đi ạ

    \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{29}{42}:\frac{1}{28}-8\)

\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{29}{42}\right):\frac{1}{28}-8\)

\(=\left(\frac{9}{42}-\frac{26}{21}+\frac{29}{42}\right):\frac{1}{28}-8\)

\(=\left(\frac{-17}{42}+\frac{29}{42}\right):\frac{1}{28}-8\)

\(=\frac{12}{42}:\frac{1}{28}-8\)

\(=\frac{2}{7}.28-8\)

\(=\frac{56}{7}-8\)

\(=8-8\)

\(=0\)

36/B
37/C
38/D
39/D
40/C

36.B
37.C
38.D
39.D
40.C

b

c

d

d

c

lm hộ mik vs mn ơiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

1: Ta có: \(x^2-16=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

2: Ta có: \(1-36x^2=0\)

\(\Leftrightarrow\left(6x-1\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

\(7,=\left(0,5a+5b\right)\left(0,25a^2-2,5ab+25b^2\right)\\ 8,=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\\ 9,=\left(5+a-b\right)\left(25-5a+5b+a^2-2ab+b^2\right)\)

\(61,\\ 1,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ 2,\Leftrightarrow\left(1-6x\right)\left(1+6x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ 3,\Leftrightarrow\left(6-x\right)\left(6+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Câu 61:

1. x² - 16 = 0

<=> x² - 4² = 0

<=> (x - 4)(x + 4) = 0

<=> \(\left[{}\begin{matrix}x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

a) \(2^4+8\left[\left(-2\right)^2:\dfrac{1}{2}\right]^0-2^{-2}.4+\left(-2\right)^2\)

\(=2^4+8.1-\dfrac{1}{4}.4+4\)

\(=16+8-1+4\)

\(=24-1+4\)

\(=23+4\)

\(=27\)

Bài 4: 
\(8^{12}\)\(-2^{33}\)-\(2^{30}\) chia hết cho 55
= \(2^{36}\)-\(2^{33}\)-\(2^{30}\)
= \(2^{30}.\left(2^6-2^3-1\right)\)
= \(2^{30}\). 55 chia hết cho 55 (đpcm)

Bài 3:
Ta có: \(5^{1000}=\left(5^2\right)^{500}=25^{500}\)
\(3^{1500}=\left(3^3\right)^{500}=27^{500}\)
Do \(25^{500}< 27^{500}\Rightarrow5^{1000}< 3^{1500}\)

1: Xét ΔMDB vuông tại D và ΔNEC vuông tại E có

BD=CE

góc MBD=góc NCE

=>ΔMDB=ΔNEC

=>DM=EN

2: DM//EN

DM=EN

=>DMEN là hình bình hành

=>I là trung điểm của MN