Bài 8: 

a) \(\left(-3,5\right):\left(-2\dfrac{3}{5}\right)=\dfrac{7}{2}:\dfrac{13}{2}=\dfrac{7}{2}\cdot\dfrac{2}{13}=\dfrac{7\cdot2}{2\cdot13}=\dfrac{7}{13}\)

b) \(\left(-\dfrac{11}{15}\right):1\dfrac{1}{10}=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}=\left(-\dfrac{11}{15}\right)\cdot\dfrac{10}{11}=\dfrac{-11\cdot10}{15\cdot11}=-\dfrac{10}{15}=-\dfrac{2}{3}\)

c) \(2\dfrac{2}{3}:\left(-3\dfrac{3}{4}\right)=\dfrac{8}{3}:-\dfrac{15}{4}=\dfrac{8}{3}\cdot-\dfrac{4}{15}=\dfrac{8\cdot4}{3\cdot15}=-\dfrac{32}{45}\)

Bài 7:

a) \(\left(-\dfrac{3}{25}\right):6=\left(-\dfrac{3}{25}\right)\cdot\dfrac{1}{6}=\dfrac{-3\cdot1}{25\cdot6}=-\dfrac{1}{50}\)

b) \(-\dfrac{5}{23}:-2=\dfrac{5}{23}\cdot\dfrac{1}{2}=\dfrac{5\cdot1}{23\cdot2}=\dfrac{5}{26}\)

c) \(\dfrac{-7}{11}:-3,5=\dfrac{7}{11}:\dfrac{7}{2}=\dfrac{7}{11}\cdot\dfrac{2}{7}=\dfrac{7\cdot2}{11\cdot7}=\dfrac{2}{11}\)

a.=-2x³y² có bậc là:5

b.=15x⁴y⁴z có bậc là:9

c.=-2/3x³y⁴ có bậc là 7

d.=-2x⁶y⁶ có bậc là:12

đợi mik đi ăn cơm xong làm cho h mik off rùi :((

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)

Câu 4: 

Số đo các góc còn lại là \(47^0;133^0;133^0\)

1B

2 bạn có chép đúng đề?

3D

4B

5A

6A

7A

8A

9B

10B

11C

12A

13C

14B

15A

16C

a: Xét ΔABC có \(\hat{A}+\hat{C}+\hat{B}=180^0\)

=>\(\hat{C}-10^0+\hat{C}+\hat{C}+10^0=180^0\)

=>\(3\cdot\hat{C}=180^0\)

=>\(\hat{C}=60^0\)

=>\(\hat{A}=60^0-10^0=50^0\)

\(\hat{B}=60^0+10^0=70^0\)

b: Xét ΔABC có \(\hat{A}+\hat{C}+\hat{B}=180^0\)

=>\(\hat{C}+\frac76\cdot\hat{C}+\frac56\cdot\hat{C}=180^0\)

=>\(3\cdot\hat{C}=180^0\)

=>\(\hat{C}=60^0\)

\(\hat{B}=\frac76\cdot60^0=70^0;\hat{A}=\frac56\cdot60^0=50^0\)

c: Đặt \(a=\hat{A};b=\hat{B};c=\hat{C}\)

\(21\cdot\hat{A}=14\cdot\hat{B}=6\cdot\hat{C}\)

=>21a=14b=6c

=>\(\frac{21a}{42}=\frac{14b}{42}=\frac{6c}{42}\)

=>\(\frac{a}{2}=\frac{b}{3}=\frac{c}{7}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{7}=\frac{a+b+c}{2+3+7}=\frac{180^0}{12}=15^0\)

=>\(\begin{cases}a=15^0\cdot2=30^0\\ b=15^0\cdot3=45^0\\ c=15^0\cdot7=105^0\end{cases}\Rightarrow\hat{A}=30^0;\hat{B}=45^0;\hat{C}=105^0\)

\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)

\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{8}{1}=8\)

Lm ơn giúp mik đii mà mik bt ơn bn đó nhiều lắm . Mik đang rất cần

Vì ( x + 23 ) . ( x - 8 ) = 0

nên ( x + 23 ) hoặc ( x - 8 ) bằng 0

x + 23 = 0                                           x - 8 = 0

x         = 0 - 23                                    x      = 0 + 8 

x         = -23                                        x      = 8

Vậy x thuộc { -23;8}

=>x+23=0

x-8=0

=> x=-23

x=8

Vạy x=-23, x=8