BĐT cần chứng minh tương đương:

\(\left(a+b\right)\left(\dfrac{a+b}{ab}\right)\ge4\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

Dấu "=" xảy ra khi và chỉ khi \(a=b\)

Áp dụng bất đẳng thức Cô-si ta có:

\((a+b)\ge 2\sqrt{ab}\)

\(\left(\dfrac1a+\dfrac1b\right)\ge 2\sqrt{\dfrac1{ab}}\)

\(\Rightarrow (a+b)\left(\dfrac1a+\dfrac1b\right) \ge 2\sqrt{ab}2\sqrt{\dfrac1{ab}}=4\) (đpcm)

Dấu \("="\) xảy ra khi \(a=b\)

Áp dụng BĐT với hai số dương ta có:

`a+b>=2sqrt{ab}`

`1/a+1/b>=2/sqrt{ab}`

`=>(a+b)(1/a+1/b)>=2sqrt{ab}. 2/sqrt{ab}=4`

Dấu "=" xảy ra khi `a=b>0`

mk thấy cm \(\dfrac{a^2+b^2}{2}\ge ab\)   thì đúng hơn

Sửa đề: \(\dfrac{a^2+b^2}{2}\ge ab\)

Ta có: \(\left(a-b\right)^2\ge0\) với mọi a, b

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge ab\)

Dấu "=" xảy ra khi a=b

Lời giải:
a. 

$\frac{a}{b}< \frac{c}{d}\Rightarrow \frac{a}{b}-\frac{c}{d}<0$

$\Rightarrow \frac{ad-bc}{bd}< 0$

$\Rightarrow ad-bc<0$ (do $bd>0$)

$\Rightarrow ad< bc$ (đpcm)

b.

$\frac{a}{b}-\frac{a+c}{b+d}=\frac{a(b+d)-b(a+c)}{b(b+d)}=\frac{ad-bc}{b(b+d)}<0$ do $ad-bc<0$ và $b(b+d)>0$

$\Rightarrow \frac{a}{b}< \frac{a+c}{b+d}$

--------

$\frac{a+c}{b+d}-\frac{c}{d}=\frac{d(a+c)-c(b+d)}{d(b+d)}=\frac{ad-bc}{d(b+d)}<0$ do $ad-bc<0$ và $d(b+d)>0$

$\Rightarrow \frac{a+c}{b+d}< \frac{c}{d}$
Ta có đpcm.

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

A>B

A>B

nếu cần mik chụp cho bạn 1 đề tương tự

a: \(A=\left(5xy-2xy+4xy\right)+3x-2y-y^2\)

\(=7xy+3x-2y-y^2\)

b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)

\(=\dfrac{-7}{8}ab^2+\dfrac{3}{8}a^2b\)

c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)

\(=7a^2b-11b^2+9c^2\)

\(A=5xy-y^2-2xy+4xy+3x-2y\)

\(A=-y^2+7xy+3x-2y\)

\(B=\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2+\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b-\dfrac{1}{2}ab^2\)

\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)

\(C=2a^2b-8b^2+5a^2b+5c^2-3b^2+4c^2\)

\(C=7a^2b-11b^2+9c^2\)

\(A=7xy-y^2+3x-2y\)

\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)
\(C=7a^2b-11b^2+9c^2\)

1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

Chọn A

a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)