5m

Ta có \(a-b=7\) \(\Rightarrow a=7+b\)

Thay vào biểu thức P ta được:

\(P=\dfrac{3\left(7+b\right)-b}{2\left(7+b\right)+7}+\dfrac{3b-\left(7+b\right)}{2b-7}\\ \Leftrightarrow P=\dfrac{21+2b}{21+2b}+\dfrac{2b-7}{2b-7}\\ \Leftrightarrow P=1+1\\ \Leftrightarrow P=2\)

     Vậy...

a, \(2^3.2^5=2^8=256\)   

    \(\left(-3\right)^9:\left(-3\right)^5=\left(-3\right)^4=81\)     

    \(\left(-6\right)^9.6^5=\left(-1\right)^9.6^9.6^5=\left(-1\right).6^{14}\\ \left(\dfrac{1}{2}\right)^5=\dfrac{1}{32}\)

b, \(\left(\dfrac{3}{5}\right)^6.\left(\dfrac{5}{3}\right)^6=\left(\dfrac{3}{5}. \dfrac{5}{3}\right)^6=1^6=1\\ \left(-\dfrac{7}{8}\right)^9:\left(\dfrac{7}{4}\right)^9=\left(-\dfrac{7}{8}:\dfrac{7}{4}\right)^9=\left(-\dfrac{1}{2}\right)^9=-\dfrac{1}{512}\\ \left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(\dfrac{1}{2}\right)^{...}\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)...\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)^6\)

c, \(\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\left(\dfrac{2}{3}\right)^4\right)^2=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^2\\ \left(\dfrac{1}{3}\right)^{12}:\left(-\dfrac{3}{9}\right)^{12}=\left(\dfrac{1}{3}.\left(-3\right)\right)^{12}=\left(-1\right)^{12}=1\\ \left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{3}\right)^{10}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)

a) Xét ΔABD và ΔACD có

AB=AC(ΔABC cân tại A)

∠∠

∠∠(AD là tia phân giác của ∠)

AD: chung

 ⇒ΔABD=ΔACD(c-g-c)

b) Ta có: ΔABD=ΔACD(cmt)

nên BD=CD(Hai cạnh tương ứng)

hay D là trung điểm của BC

Xét ΔABC có 

AD là đường trung tuyến ứng với cạnh BC(cmt)

G là trọng tâm của ΔABC(gt)

⇒ A,G,D thẳng hàng(đpcm)

xem đc r

1.A picture is being painted by my daughter.

2.Beef isn't being eaten by Sam.

3.Soccer is played in the yard. 

4.A novel is being read in class now.

5.Pottery is made in his friend's room by him everyday.

???  Sao tui chẳng thấy j nhể

@AnkTrần không có gì đâu ạ 

@AnkTrần tắc chỗ nào bảo mik nhé

Ta có:

\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)

Ta thấy:

\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)

Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)

\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)

 Vậy \(x=-100\)

@Jen vậy đợi mik xem đã