1) Tính góc ∠E1

Ta có d’//d” (gt)

⇒ ∠C = ∠E1 ( So le trong)

⇒ ∠E1 = 600 vì ∠C = 600

2) Tính ∠G3

Ta có d’//d”

⇒ ∠G2 = ∠D (Đồng vị)

⇒ ∠G1 = 1100

3) Tính ∠G3

Vì ∠G2 + ∠G3 = 1800 (kề bù)

⇒ ∠G3 = 700

4) Tính ∠D4

∠D4 = ∠D (Đối đỉnh)

⇒  ∠D4 = 1100

5) Tính ∠A5

Ta có d//d”

⇒ ∠A5 = ∠ E1 (Đồng vị)

⇒ ∠A5 = 600

6) Tính ∠B6

Ta có d//d”

⇒ ∠B6 = ∠G3 (Đồng vị)

⇒ ∠B6 = 700

GT : a//c ; b//c
KL : a//c

hok tốt

^_^

1.(25¹⁵.4¹⁵)/(5¹⁷.20¹⁶)=(5³⁰.2³⁰)/(5¹⁷.5¹⁶.2³²)=1/(5³.2²)=1/500

2.a,-x/2=8/-x=>-x.(-x)=2*8 =>x^2=16=(-4)^2=4^2

=>x=4 hoặc x=-4

b,(3/4)^2x/(2/5)¹⁰=(15/8)¹⁰

(3/4)^2x=(2/5*15/8)¹⁰

(3/4)^2x=(3/4)¹⁰

2x=10

x=5

1)  \(\frac{25^{15}\cdot4^{15}}{5^{17}\cdot20^{16}}\)

\(=\frac{5^{30}\cdot2^{30}}{5^{33}\cdot2^{32}}\)

\(=\frac{1}{5^3\cdot2^2}\)

\(=\frac{1}{500}\)

2) 

a) \(\frac{-x}{2}=\frac{8}{-x}\)

\(\Rightarrow\left(-x\right)\left(-x\right)=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\left\{\pm4\right\}\)

nany???

ai cho copy bài làm của tui

\(\left|2x+3\right|=x+2\)(1)

- Nếu \(2x+3\ge0\Rightarrow x\ge\frac{-3}{2}\)thì (1) trở thành: \(2x+3=x+2\Rightarrow x=-1\) (nhận)

=Nếu \(2x+3< 0\Rightarrow x< \frac{-3}{2}\) thì (1) trở thành: \(2x+3=-\left(x+2\right)\Rightarrow2x+3=-x-2\Rightarrow3x=-5\Rightarrow x=\frac{-5}{3}\) (nhận)

Vậy \(x=\left\{\frac{-5}{3};-1\right\}\)

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\left(\frac{x+4}{2000}+\frac{2000}{2000}\right)+\left(\frac{x+3}{2001}+\frac{2001}{2001}\right)=\left(\frac{x+2}{2002}+\frac{2002}{2002}\right)+\left(\frac{x+1}{2003}+\frac{2003}{2003}\right)\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(x+2004\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Ta thấy   \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)

nên  \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\ne0\)

Do đó: x + 2004 = 0 => x = -2004

  Vậy x = -2004

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

Nên x + 2004 = 0

=> x = -2004

Vậy x = -2004