Ta có:

\(\dfrac{2a+b}{a+b}+\dfrac{2c+d}{c+d}+\dfrac{2b+c}{b+c}+\dfrac{2d+a}{d+a}=6\)

⇔ \(\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)=2\)

⇔ \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)

⇔ \(\left(1-\dfrac{a}{a+b}\right)-\dfrac{b}{b+c}+\left(1-\dfrac{c}{c+d}\right)-\dfrac{d}{d+a}=0\)

⇔ \(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)

⇔ \(\dfrac{b\left(b+c\right)-b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(d+a\right)-d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{d\left(c-a\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\left(c-a\right)\left(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right)=0\)

⇒ \(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}=0\)         \(\left(a\ne c\right)\)

⇒ \(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)

⇔ \(\left(bc+bd\right)\left(d+a\right)-\left(ad+bd\right)\left(b+c\right)=0\)

⇔ \(bcd+abc+bd^2+abd-abd-acd-b^2d-bcd=0\)

⇔ \(abc+bd^2-acd-b^2d=0\)

⇔ \(ac\left(b-d\right)-bd\left(b-d\right)=0\)

⇔ \(\left(b-d\right)\left(ac-bd\right)=0\)

⇒ \(ac-bd=0\)       \(\left(b\ne d\right)\)

⇔ \(ac=bd\)

Khi đó:

\(A=abcd=\left(ac\right)^2\)

⇒ \(ĐPCM\)

 1.  I (go) _____went______ to the cinema three times last week.

     2.  I was angry because they (be) _____were______ late.

     3.  ______Did_____ you (attend) _____attend______ the extra class last night.

     4.  Last year I (live) ______lived______ in London.

     5.  My best friend (go) _____went______ abroad five days ago.

     6.  The weather (be) ______was_____ good when we were on holiday.

     7.  I (meet) _____met______ Tom and Ann at the airport a few weeks ago.

     8.  I (buy) ______bought_____ that novel on Monday.

     9.  Liza (get) _____got______ a good mark in the final exam.

     10. _____Did______ you (go) _____go______ out yesterday evening?

Đây ạ...

https://hoc24.vn/cau-hoi/ho-em-bai-1-va-bai-2-voi-a-em-cam-on.400015920632

a)

\(\text{Δ A'B'C' ∼ Δ ABC}\) theo tỉ số đồng dạng k = \(\dfrac{2}{5}\)

⇒ \(\dfrac{A'B'}{AB}=\dfrac{B'C'}{BC}=\dfrac{A'C'}{AC}=k=\dfrac{2}{5}\)              (1)

Áp dúng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{A'B'}{AB}=\dfrac{B'C'}{BC}=\dfrac{A'C'}{AC}=\dfrac{A'B'+B'C'+A'C'}{AB+BC+AC}=\dfrac{C_{A'B'C'}}{C_{ABC}}\)                 (2)

Từ (1) và (2) ⇒ \(\dfrac{C_{A'B'C'}}{C_{ABC}}=\dfrac{2}{5}\)           (*)

b)

Theo đề ra, ta có:

\(C_{ABC}-C_{A'B'C'}=30\left(cm\right)\)

⇒ \(C_{ABC}=30+C_{A'B'C'}\)      (**)

Thay (**) vào (*), ta được:

\(\dfrac{C_{A'B'C'}}{30+C_{A'B'C'}}=\dfrac{2}{5}\)

⇒ \(5C_{A'B'C'}=60+2C_{A'B'C'}\)

⇔ \(3C_{A'B'C'}=60\)

⇒ \(C_{A'B'C'}=20\)     (cm)

⇒ \(C_{ABC}=30+20=50\)   (cm)

a)

\(\text{Δ A'B'C' ∼ Δ ABC}\) theo tỉ số đồng dạng k = \(\dfrac{4}{7}\)

⇒ \(\dfrac{A'B'}{AB}=\dfrac{B'C'}{BC}=\dfrac{A'C'}{AC}=k=\dfrac{4}{7}\)              (1)

Áp dúng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{A'B'}{AB}=\dfrac{B'C'}{BC}=\dfrac{A'C'}{AC}=\dfrac{A'B'+B'C'+A'C'}{AB+BC+AC}=\dfrac{C_{A'B'C'}}{C_{ABC}}\)                 (2)

Từ (1) và (2) ⇒ \(\dfrac{C_{A'B'C'}}{C_{ABC}}=\dfrac{4}{7}\)           (*)

b)

Theo đề ra, ta có:

\(C_{ABC}-C_{A'B'C'}=90\left(dm\right)\)

⇒ \(C_{ABC}=90+C_{A'B'C'}\)      (**)

Thay (**) vào (*), ta được:

\(\dfrac{C_{A'B'C'}}{90+C_{A'B'C'}}=\dfrac{4}{7}\)

⇒ \(7C_{A'B'C'}=360+4C_{A'B'C'}\)

⇔ \(3C_{A'B'C'}=360\)

⇒ \(C_{A'B'C'}=120\)     (dm)

⇒ \(C_{ABC}=120+90=210\)   (dm)