\(a,157.17-157.7=157\left(17-7\right)=1570\)

\(b,5.\left(-3+2\right)-7.\left(5-4\right)=-5-7=-12\)

\(c,17.\left(-84\right)+17\left(-16\right)=17\left[\left(-84\right)+\left(-16\right)\right]=-1700\)

\(d,-145.\left(13-57\right)+57.\left(10-145\right)=6380-7695=-1315\)

\(e,199.\left(15-17\right)-199.\left(-17+5\right)=199.10=1990\)

Nhìn nó rối ghê á

1: =-15+10-35+28=-32

rối thật đó

3: \(=20-12-8+12=20-8=12\)

5: \(=-18-42-21-35=-116\)

3: \(=-15+18-12+8=-27+26=-1\)

2: \(=-12+21-15+10=9-5=4\)

1: =-15+10-35+28=-12

3: =20-12-8+12=12

2) -3(4 - 7) + 5(-3 + 2)

= -3.4 + 3.7 - 5.3 + 5.2

= -12 + 21 -15 + 10

= 31 - 27

= 4

4) -5(2 - 7) + 4(2 - 5)

= -5.2 + 5.7 + 4.2 - 4.5

= -10 + 35 + 8 - 20

= 38 - 30

= 8

5) 6(-3 - 7) - 7(3 + 5)

= -6.3 - 6.7 - 7.3 - 7.5

= -18 - 42 - 21 - 35

= -116

6) 3(-5 + 6) - 4(3 - 2)

= -3.5 + 3.6 - 4.3 + 4.2

= -15 + 18 - 12 + 8

= 26 - 27

= -1

\(a^7-a^5+2a^3+2a^2\\ =a^5\left(a^2-1\right)+2a^2\left(a+1\right)\\ =a^5\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)\\ =a^2\left(a+1\right)\left(a^4-a^3+2\right)\)

\(a^7-a^5+2a^3+2a^2=a^5\left(a^2-1\right)+2a^2\left(a+1\right)\)

\(=\left(a+1\right)\left[a^5\left(a-1\right)+2a^3\right]\)

\(=a^3\left(a+1\right)\left[a^2\left(a-1\right)+2\right]\)

\(=a^3\left(a+1\right)\left(a^3-a^2-2\right)\)

a)

 \(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{{ - 5}}{{56}} + \frac{{ - 11}}{{56}} = \frac{{ - 16}}{{56}} = \frac{{ - 2}}{7}\end{array}\)

b)

\(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{1}{7}.[(\frac{{ - 5}}{8}) + (\frac{{ - 11}}{8})]\\ = \frac{1}{7}.\frac{{ - 16}}{8}\\ = \frac{1}{7}.( - 2)\\ = \frac{{ - 2}}{7}\end{array}\)

a.

\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)

\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)

b.

\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)

c.

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)

\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)

\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)

\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)