\(2+2=4\)

\(4+4=8\)

\(6+6=12\)

\(8+8=16\)

\(10+10=20\)

       k mk nha !

a) \(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)( đảo trung tỉ )

\(\Rightarrow\frac{c}{a}=\frac{d}{b}\)( nghịch đảo )

\(\Rightarrow\frac{c}{a}+1=\frac{d}{b}+1\)

\(\Rightarrow\frac{c}{a}+\frac{a}{a}=\frac{d}{b}+\frac{b}{b}\)

\(\Rightarrow\frac{c+a}{a}=\frac{d+b}{b}\)

\(\Rightarrow\frac{a}{a+c}=\frac{b}{b+d}\)(nghịch đảo)

b) Chứng minh tương tự phần a; ta lại có:

\(\frac{c}{a}=\frac{d}{b}\)

\(\Rightarrow1-\frac{c}{a}=1-\frac{d}{c}\)

\(\Rightarrow\frac{a}{a}-\frac{c}{a}=\frac{b}{b}-\frac{d}{b}\)

\(\Rightarrow\frac{a-c}{a}=\frac{b-d}{b}\)

\(\Rightarrow\frac{a}{a-c}=\frac{b}{b-d}\)(nghịch đảo)

ex 1: khoang tron vao dap an dung

1. you should get ___ to check cracks in the pipes in your house.

A. worker B. plumber C. driver D. mechanic

2. i'm responsible for cooking dinner as my mother usually works ___.

A. lately B. early C. later D. late

3. Ms. Mai asked me how she could ____ household chores equally in her family.

A. make B. divide C. give D. contribute

4. in my family, my father always takes charge of doing the ____ lifting.

A. strong B. hard C. heavy D. huge

5. my mother told me to do the ____ yesterday but i forgot about it since i had much homwork to finish.

A. wash-up B. laundry C. childcare D. exercises

6. ___ the rubbish in the early morning is a part of my daily routine.

A. taking out B. coming out C. pulling out D. bringing out

m.n giup mk cau nay aj. thank m.n

áp dụng BĐT Cauchy ngược dấu:

\(x+y=3+\sqrt{xy}\le3+\frac{x+y}{2}\)

\(=>\frac{x+y}{2}\le3=>x+y\le6\left(1\right)\)

\(4=\sqrt{x+1}+\sqrt{y+1}\)

=> \(8=\sqrt{4\left(x+1\right)}+\sqrt{4\left(y+1\right)}\le\frac{4\left(x+1\right)}{2}+\frac{4\left(y+1\right)}{2}=>x+y\ge6\left(2\right)\)

từ (1) và (2) => x+ y = 6

<=> \(\left\{{}\begin{matrix}x=y\\4=\left(x+1\right)=>x=y=3\\4=\left(y+1\right)\end{matrix}\right.\)

vậy hpt có No (x; y) = (3;3)

\(\left(\frac{15\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)+4\left(\sqrt{6}+1\right)\left(3-\sqrt{6}\right)-12\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)}\right)=\left(\frac{15\left(3\sqrt{6}-6-6+2\sqrt{6}\right)+4\left(3\sqrt{6}-6+3-\sqrt{6}\right)-12\left(6-2\sqrt{6}+\sqrt{6}-2\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)}\right)=(\frac{\left(15\left(5\sqrt{6}-12\right)+4\left(2\sqrt{6}-3\right)-12\left(4-\sqrt{6}\right)\right)}{18-7\sqrt{6}})=\left(\frac{75\sqrt{6}-180+8\sqrt{6}-12-48+12\sqrt{6}}{18-7\sqrt{6}}\right)=\left(\frac{95\sqrt{6}-240}{18-7\sqrt{6}}\right)=\left(\frac{95\sqrt{6}-240}{7\sqrt{6}-18}\right)\)