chinh lại câu c.

x+2/5=2-3x

x+3x=2-2/5

4x=8/5

x=2/5

a. /2x-1/=/x+2/

2x-1= x+2 hoặc 2x-1=-x-2

x=3 hoặc 3x=-1

x=3 hoặc x=-1/3

b. 2.\(3^x\).\(3^2\)=18

\(3^x\).18=18

\(3^x\)=1

\(3^x\)=\(3^0\)

x=0

c.x+2/5=2-3x

x+3x=2-2/5

4x=3/5

x=3/20

.x+2/5=2-3x/3

x+2/5=2-x

x+x=2-2/5

2x=8/5

x=4/5

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)

\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)

Tương tự: \(\dfrac{b}{\sqrt{b^2+a+c}}\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\) ; \(\dfrac{c}{\sqrt{c^2+b+a}}\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)

Cộng vế:

\(P\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)

Lại có:

\(a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}\)

\(=\sqrt{a}.\sqrt{a+ab+ac}+\sqrt{b}.\sqrt{b+bc+ab}+\sqrt{c}.\sqrt{c+ac+bc}\)

\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+2bc+2ca\right)}\)

\(\Rightarrow P\le\dfrac{\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+bc+ca\right)}}{a+b+c}=\sqrt{\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}}\)

Do đó ta chỉ cần chứng minh:

\(\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}\le3\Leftrightarrow a+b+c\ge ab+bc+ca\)

Thật vậy:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow a+b+c\ge ab+bc+ca\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

C/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\) (*)

Thật vậy , (*) \(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow ab+bc+ac+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ac\right)+4\left(a+b+c\right)+8\)

\(\Leftrightarrow ab+bc+ac+abc=4\) (Đ)

=> (*) đúng ( đpcm ) 

b. Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k => a=bk; c=dk

Vế trái =\(\frac{a^2}{b^2}\)=\(\frac{b^2k^2}{b^2}\)=\(k^2\)(1)

Vế phải =\(\frac{a^2-ac}{b^2-bd}\)=\(\frac{b^2k^2-bk.dk}{b^2-bd}\)=\(\frac{k^2\left(b^2-bd\right)}{b^2-bd}\)=\(k^2\)(2)

từ (1) và (2) ta có\(\frac{a^2}{b^2}\)=\(\frac{a^2-ac}{b^2-bd}\)

b.Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)=k => a=bk; c=dk

Vế trái =\(\frac{5a+5b}{5b}\)=\(\frac{5bk+5b}{5b}\)=\(\frac{5b\left(k+1\right)}{5b}\)=k+1(1)

Vế phải =\(\frac{c^2+cd}{cd}\)=\(\frac{d^2.k^2+d^2.k}{d^2.k}\)=\(\frac{d^2.k\left(k+1\right)}{d^2.k}\)=k+1(2)

từ (1) và (2) ta có\(\frac{5a+5b}{5b}\)=\(\frac{c^2+cd}{cd}\)

mình viết thiếu đề:

Cho tỉ lệ thức sau: a/b = c/d chứng minh:

\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)

Tương tự:

\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)

Cộng vế:

\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)