\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)

\(\Leftrightarrow\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10},Đkxđ:x\ne0,6,-10\)

\(\Leftrightarrow\frac{60}{x}-\frac{30}{x-6}-\frac{30}{x+10}=0\)

\(\Leftrightarrow\frac{60\left(x-6\right)\left(x+10\right)-30x\left(x+10\right)=30\left(x-6\right)}{x\left(x-6\right)\left(x+10\right)}\)

\(\Leftrightarrow\frac{\left(60x-360\right)\left(x+10\right)-30x^2-300x-30x^2+180x}{x\left(x-6\right)\left(x+10\right)}\)

\(\Leftrightarrow\frac{60x^2+600x-360x-3600-30x^2-300x-30x^2+180}{x\left(x-6\right)\left(x=10\right)}=0\)

\(\Leftrightarrow\frac{120x-3600}{x\left(x-6\right)\left(x+10\right)}=0\)

\(\Leftrightarrow120x-3600=0\)

\(\Leftrightarrow120x=3600\)

\(\Leftrightarrow x=30;x\ne0;x\ne6,x\ne-10\)

1/

\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)

\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)

2/

\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)

\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)

3/

\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)

\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)

\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)

\(=-1+2cos2a+1-cos2a=cos2a\)

30 theo tui nghĩ là đúng

=>\(\frac{30x\left(x-6\right)}{x\left(x+10\right)\left(x-6\right)}+\frac{30x\left(x+10\right)}{x\left(x+10\right)\left(x-6\right)}=\frac{60\left(x+10\right)\left(x-6\right)}{x\left(x-6\left(x+10\right)\right)}\)

=>30x²-180x+30x²+300x=60x²-360x+600x-3600

=>60x²+120x=60x²+240x-3600

=>-120x=-3600

=>x=30

nhớ k mk........Đúng 100%

Ta có:\(\frac{60}{x}\)+    \(\frac{30}{y}\)=     4

Vậy ta sẽ chuyển hai phân số đã cho về số tự nhiên là 2 + 2 hoặc là 3+1

Cách 1 :Nếu muốn thành 2+2 thì tử sẽ gấp 2 lần mẫu vậy x là:60:2=30

y là:30:2=15

Cách 1: ( chọn )

Cách 2:Nếu muốn thành 3+1 thì phân số thứ 1 phải có tử gấp 3 lần mẫu và ở phân số thứ 2 phải có tử bằng mẫu vậy x là:60:3=20

y là:30:1=30

Cách 2: ( chọn )

\(\Rightarrow\)x là:30 hoặc 20

         y là:15 hoặc 30
 

Ta có : \(\frac{60}{x}+\frac{30}{y}=4\Rightarrow\frac{60}{x}+\frac{60}{2y}=4\Rightarrow60\left(\frac{1}{x}+\frac{1}{2y}\right)=4\Rightarrow\frac{1}{x}+\frac{1}{2y}=\frac{1}{15}\Rightarrow\frac{2y+x}{2xy}=\frac{1}{15}\)

=> 15(2y + x) = 2xy

=> 30y + 15x = 2xy

=> 30y + 15x - 2xy = 0

=> 30y + x(15 - 2y) = 0

=> 225 - 30y + x(15 - 2y) = 225

=> 15(15 - 2y) + x(15 - 2y) = 225

=> (15 + x)(15 - 2y) = 225 

Bạn tự lập bảng xét các trường hợp 

A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)

B=0

\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)

\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)

A=\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)

A=\(\left(\frac{1}{7}-\frac{1}{8}\right).\left(\frac{1}{8}-\frac{1}{9}\right).\left(\frac{1}{9}-\frac{1}{10}\right).\left(\frac{1}{10}-\frac{1}{11}\right).\left(\frac{1}{11}-\frac{1}{12}\right).\left(\frac{1}{12}-\frac{1}{13}\right).\left(\frac{1}{13}-\frac{1}{14}\right)\)

A=\(\frac{1}{7}-\frac{1}{14}\)

A=\(\frac{1}{14}\)

a) \(M = \sin {45^o}.\cos {45^o} + \sin {30^o}\)

Ta có: \(\left\{ \begin{array}{l}\sin {45^o} = \cos {45^o} = \frac{{\sqrt 2 }}{2};\;\\\sin {30^o} = \frac{1}{2}\end{array} \right.\)

Thay vào M, ta được: \(M = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2} = \frac{2}{4} + \frac{1}{2} = 1\)

b) \(N = \sin {60^o}.\cos {30^o} + \frac{1}{2}.\sin {45^o}.\cos {45^o}\)

Ta có: \(\sin {60^o} = \frac{{\sqrt 3 }}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\sin {45^o} = \frac{{\sqrt 2 }}{2};\, \cos {45^o}= \frac{{\sqrt 2 }}{2}\)

Thay vào N, ta được: \(N = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{3}{4} + \frac{1}{4} = 1\)

c) \(P = 1 + {\tan ^2}{60^o}\)

Ta có: \(\tan {60^o} = \sqrt 3 \)

Thay vào P, ta được: \(Q = 1 + {\left( {\sqrt 3 } \right)^2} = 4.\)

d) \(Q = \frac{1}{{{{\sin }^2}{{120}^o}}} - {\cot ^2}{120^o}.\)

Ta có: \(\sin {120^o} = \frac{{\sqrt 3 }}{2};\;\;\cot {120^o} = \frac{{ - 1}}{{\sqrt 3 }}\)

Thay vào P, ta được: \(Q = \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \;{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\frac{3}{4}}} - \;\frac{1}{3} = \;\frac{4}{3} - \;\frac{1}{3} = 1.\)

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

\(\frac{10}{25}=\frac{10:5}{25:5}=\frac{2}{5}\)

\(\frac{5}{20}=\frac{5:5}{20:5}=\frac{1}{4}\)

\(\frac{6}{30}=\frac{6:6}{30:6}=\frac{1}{5}\)

\(\frac{60}{20}=\frac{60:20}{20:20}=3\)