Bài 13:

góc A=180-80-30=70 độ

=>góc BAD=góc CAD=70/2=35 độ

góc ADC=80+35=115 độ

góc ADB=180-115=65 độ

Bài 14: 
Xét ΔABC vuông tại A 
-> \(\widehat{B}\)\(+ \widehat{C}=90^o\)

Mà \(\widehat{B}=\widehat{C}\)
=> \(2\widehat{B}=90^o\)
=> \(\widehat{B}=45^o\)

14.

\(\dfrac{1-cosa}{sina}=\dfrac{sina\left(1-cosa\right)}{sin^2a}=\dfrac{sina\left(1-cosa\right)}{1-cos^2a}=\dfrac{sin\left(1-cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\dfrac{sina}{1+cosa}\)

Câu b đề bài sai, đẳng thức đúng phải là:  \(1+tan^2a=\dfrac{1}{cos^2a}\)

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

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

\(\dfrac{sin^4a-cos^4a}{sina+cosa}=\dfrac{\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)}{sina+cosa}=\dfrac{sin^2a-cos^2a}{sina+cosa}=\dfrac{\left(sina+cosa\right)\left(sina-cosa\right)}{sina+cosa}\)

\(=sina-cosa\)

13.

b. Chia cả tử và mẫu cho sinB:

\(N=\dfrac{\dfrac{4cosB}{sinB}+\dfrac{2sinB}{sinB}}{\dfrac{cossB}{sinB}-\dfrac{3sinB}{sinB}}=\dfrac{4cotB+2}{cotB-3}=\dfrac{4.\dfrac{3}{2}+2}{\dfrac{3}{2}-3}=-\dfrac{16}{3}\)

c. Chia cả tử và mẫu cho \(cos^3B\)

\(M=\dfrac{\dfrac{sin^3B}{cos^3B}-\dfrac{cos^3B}{cos^3B}}{\dfrac{sin^3B}{cos^3B}+\dfrac{cos^3B}{cos^3B}}=\dfrac{tan^3B-1}{tan^3B+1}=\dfrac{3^3-1}{3^3+1}=\dfrac{13}{14}\)

Bài 11:

Gọi số học sinh giỏi 4 khối lần lượt là $a,b,c,d$ (em) 

Theo bài ra ta có: $a+b+c-d=168$ và $\frac{a}{13}=\frac{b}{12}=\frac{c}{14}=\frac{d}{15}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a}{13}=\frac{b}{12}=\frac{c}{14}=\frac{d}{15}=\frac{a+b+c-d}{13+12+14-15}=\frac{168}{24}=7$

$\Rightarrow a=13.7=91; b=12.7=84; c=14.7=98; d=15.7=105$

Bài 12:

Gọi số học sinh ba khối lần lượt là $a,b,c$ (học sinh).

Theo bài ra ta có: $\frac{a}{10}=\frac{b}{9}=\frac{c}{8}$ và $a-b=50$

Áp dụng TCDTSBN:

$\frac{a}{10}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{10-9}=\frac{50}{1}=50$

$\Rightarrow a=50.10=500; b=50.9=450; c=50.8=400$ (hs)

13:

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}sin\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{pi}{33}\right)\cdot cos\left(\dfrac{2pi}{33}\right)\cdot cos\left(\dfrac{4pi}{33}\right)\cdot cos\left(\dfrac{8pi}{33}\right)\cdot cos\left(\dfrac{16pi}{33}\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{2}\cdot sin\dfrac{2}{33}pi\cdot cos\left(\dfrac{2}{33}pi\right)cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{4}\cdot sin\dfrac{4}{33}pi\cdot cos\left(\dfrac{4}{33}pi\right)\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{8}\cdot sin\dfrac{8}{33}pi\cdot cos\left(\dfrac{8}{33}pi\right)\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{33}\right)}\cdot\dfrac{1}{16}\cdot sin\dfrac{16}{33}pi\cdot cos\left(\dfrac{16}{33}pi\right)\)

\(=\dfrac{1}{sin\left(\dfrac{pi}{3}\right)}\cdot\dfrac{1}{32}\cdot sin\dfrac{32}{33}pi\)

=1/32

10:

\(=\dfrac{1}{2}\left[cos100+cos60\right]+\dfrac{1}{2}\cdot\left[cos100+cos20\right]\)

=cos100+1/2*cos20+1/4

6:

sin6*cos12*cos24*cos48

=1/cos6*cos6*sin6*cos12*cos24*cos48

=1/cos6*1/2*sin12*cos12*cos24*cos48
=1/cos6*1/4*sin24*cos24*cos48

=1/cos6*1/8*sin48*cos48

=1/cos6*1/16*sin96

=1/16

4/1 x 13/15

= 52/15

52/15

11.

\(=\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(2-\sqrt{x})(\sqrt{x}+3)}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{x-9}{(2-\sqrt{x})(\sqrt{x}+3)}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{\sqrt{x}-2}{3+\sqrt{x}}\)

12.

\(=\frac{(3-\sqrt{x})(3\sqrt{x}-2)+(5\sqrt{x}+7)(3\sqrt{x}+4)}{(5\sqrt{x}+7)(3\sqrt{x}-2)}-\frac{42\sqrt{x}+34}{(5\sqrt{x}+7)(3\sqrt{x}-2)}\) 

\(=\frac{12x+52\sqrt{x}+22}{(5\sqrt{x}+7)(3\sqrt{x}-2)}-\frac{42\sqrt{x}+34}{(5\sqrt{x}+7)(3\sqrt{x}-2)}\)

\(=\frac{12x+10\sqrt{x}-12}{(5\sqrt{x}+7)(3\sqrt{x}-2)}=\frac{2(3\sqrt{x}-2)(2\sqrt{x}+3)}{(5\sqrt{x}+7)(3\sqrt{x}-2)}=\frac{2(2\sqrt{x}+3)}{5\sqrt{x}+7}\)

Mình cần gấp ạ

\(13,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+12-3\sqrt{3}\\ =\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\\ 14,=\dfrac{12\left(4+\sqrt{10}\right)}{6}-3\sqrt{10}+\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\\ =8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\\ 15,=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}-3}\)

\(16,=\dfrac{x+2\sqrt{x}-3-x+3\sqrt{x}-4\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ 17,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

(-23)+13+(-17)+57

=[(-23)+13]+[(-17)+57]

=(-10)+40

=30

hok tốt

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Tính cách nhanh hay tính bth cx ra kq là 30 mak

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^^