B=2*C=2*2D=4*D

A=2*B=2*4D=8*D

Xét tứ giác ABCD có

góc A+góc B+góc C+góc D=360 độ

=>8*góc D+4*góc D+2*góc D+góc D=360 độ

=>góc D=24 độ

góc C=2*24=48 độ

góc B=4*24=96 độ

góc A=8*24=192 độ

b: \(sin^2b+cos^2b=1\)

=>\(sin^2b=1-\dfrac{1}{5}=\dfrac{4}{5}\)

=>\(sinb=\dfrac{2}{\sqrt{5}}\) hoặc \(sinb=-\dfrac{2}{\sqrt{5}}\)

TH1: \(sinb=\dfrac{2}{\sqrt{5}}\)

\(tanb=\dfrac{2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=2\)

cot b=1/tanb=1/2

TH2: \(sinb=-\dfrac{2}{\sqrt{5}}\)

\(tanb=\dfrac{-2}{\sqrt{5}}:\dfrac{1}{\sqrt{5}}=-2\)

cot b=1/tan b=-1/2

c: \(1+cot^2y=\dfrac{1}{sin^2y}\)

=>\(\dfrac{1}{sin^2y}=1+2=3\)

=>\(sin^2y=\dfrac{1}{3}\)

=>\(siny=\dfrac{1}{\sqrt{3}}\) hoặc \(siny=-\dfrac{1}{\sqrt{3}}\)

TH1: \(siny=\dfrac{1}{\sqrt{3}}\)

\(coty=\dfrac{cosy}{siny}\)

=>\(cosy=\dfrac{1}{\sqrt{3}}\cdot\left(-\sqrt{2}\right)=\dfrac{-\sqrt{2}}{\sqrt{3}}\)

\(tany=\dfrac{1}{coty}=\dfrac{-1}{\sqrt{2}}\)

TH2: \(siny=-\dfrac{1}{\sqrt{3}}\)

\(cosy=coty\cdot siny=\left(-\sqrt{2}\right)\cdot\dfrac{-1}{\sqrt{3}}=\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}\)

$tany=\frac{1}{coty}=\frac{-1}{\sqrt{2}}$

a: =>2x^3=58-4=54

=>x^3=27

=>x=3

b; =>(5-x)^5=2^5

=>5-x=2

=>x=3

c: =>(5x-6)^3=4^3

=>5x-6=4

=>5x=10

=>x=2

d: (3x)^3=(2x+1)^3

=>3x=2x+1

=>x=1

1=>2x³=54
=>x³=27  =>x=3
2=>(5-x)⁵=2⁵
=>5-x=2
=>x=3
3=>(5x-6)³=4³
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1

Bài 1 :

a) \(Cos30^o=Cos\left(2.15^o\right)=2cos^215^o-1\)

\(\Rightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\sqrt[]{3}+2}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)

\(\Rightarrow cos15^o=\dfrac{2\sqrt[]{\sqrt[]{3}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{4\sqrt[]{3}+8}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6+2.2\sqrt[]{2}\sqrt[]{6}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\left(\sqrt[]{6}+\sqrt[]{2}\right)^2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6}+\sqrt[]{2}^{ }}{4}\left(dpcm\right)\)

a)

 Dựng tam giác ABC vuông tại A với \(\widehat{C}=15^o\). Trên đoạn thẳng AC lấy điểm D sao cho \(\widehat{CBD}=15^o\). Không mất tính tổng quát, ta chuẩn hóa \(AB=1\). \(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{AB}{cos60^o}=2\\AD=AB.tan60^o=\sqrt{3}\end{matrix}\right.\)

 Dễ thấy tam giác DBC cân tại D \(\Rightarrow BD=CD=2\) \(\Rightarrow AC=AD+DC=2+\sqrt{3}\)

  \(\Rightarrow tanC=\dfrac{AB}{AC}=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\) 

\(\Rightarrow\dfrac{sinC}{cosC}=2-\sqrt{3}\)

\(\Rightarrow sinC=\left(2-\sqrt{3}\right)cosC\)

Mà \(sin^2C+cos^2C=1\)

\(\Rightarrow\left(7-4\sqrt{3}\right)cos^2C+cos^2C=1\)

\(\Leftrightarrow\left(8-4\sqrt{3}\right)cos^2C=1\)

\(\Leftrightarrow cos^2C=\dfrac{1}{8-4\sqrt{3}}=\dfrac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow cosC=\sqrt{\dfrac{2+\sqrt{3}}{4}}\) \(=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{8+4\sqrt{3}}}{4}\) \(=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) 

\(\Rightarrow cos15^o=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

ĐKXĐ: a>0; a<>1

a: \(A=\left(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{a-1}\cdot\dfrac{1}{\sqrt{a}}=\dfrac{2\sqrt{a}}{\sqrt{a}\left(a-1\right)}\)

\(=\dfrac{2}{a-1}\)

b: Để A là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)

=>\(a\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được:

\(a\in\left\{2;3\right\}\)

Khi a=2 thì \(A=\dfrac{2}{2-1}=2\)

Khi a=3 thì \(A=\dfrac{2}{3-1}=\dfrac{2}{2}=1\)

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

\(=\sqrt{5}-2-\sqrt{5}-2\)

=-4

2: 

a: \(B=\dfrac{2x+8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{8}{\sqrt{x}-4}\)

\(=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)

b: B là số nguyên

=>\(3\sqrt{x}⋮\sqrt{x}+1\)

=>\(3\sqrt{x}+3-3⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;-1;3;-3\right\}\)

=>\(\sqrt{x}+1\in\left\{1;3\right\}\)

=>\(x\in\left\{0;4\right\}\)

a: =>4(x-3)=49-1=48

=>x-3=12

=>x=15

b: =>123-5(x+4)=38

=>5(x+4)=123-38=85

=>x+4=17

=>x=13

c: =>2x-138=9*8=72

=>2x=72+138=210

=>x=105

\(Q=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

xin lũi nha mình cũng có bài y như này ik cứu tui dc ko