Bài 1:

1) Ta có: a<b

⇔a+5<b+5

2) Ta có: a<b

⇔a-7<b-7

3) Ta có: a<b

⇔6a<6b

4) Ta có: a<b

⇔3a<3b

hay 3a+1<3b+1

5) Ta có: a<b

⇔2a<2b

⇔-2a>-2b

hay -2a-5>-2b-5

Bài 2:

1) Ta có: a+5<b+5

⇔a<b

2) Ta có: -3a>-3b

⇔a>b

Bài 1:

a). Ta có: a < b

=> -6a > -6b

mà 3 > 1

=> \(3-6a>1-6b\)

b)

Ta có: a < b

=> a - 2 < b - 2

=> \(7\left(a-2\right)< 7\left(b-2\right)\)

c)

Ta có: a < b

=> -2a > -2b

=> 1 - 2a > 1 - 2b

\(\Rightarrow\dfrac{1-2a}{3}>\dfrac{1-2b}{3}\)

Bài 2:

a) Ta có:

a+23<b+23

\(\Leftrightarrow a< b\)

b) Ta có:

\(-12a>-12b\)

\(\Leftrightarrow a< b\)

c) Ta có:

\(5a-6\ge5b-6\)

\(a\ge b\)

d) Ta có:

\(\dfrac{-2a+3}{5}\le\dfrac{-2b+3}{5}\)

\(\Leftrightarrow-2a+3\le-2b+3\)

\(\Leftrightarrow a\ge b\)

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

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

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

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)

a: a>b

=>3a>3b

=>3a+5>3b+5

b: a>b

=>2a>2b

=>2a-3>2b-3>2b-4

a) 3a + 5 > 3b + 5;          b) 2a - 3 > 2b - 4

1.

a. -3a - 1 + 1 > -3b - 1 + 1 (cộng cả 2 vế cho 1)

  -3a . \(\left(\dfrac{-1}{3}\right)\) <  -3b . \(\left(\dfrac{-1}{3}\right)\) (nhân cả vế cho \(\dfrac{-1}{3}\) )

         a < b

b. 4a + 3 + (- 3) < 4b + 3 +(- 3) (cộng cả 2 vế cho -3)

   4a . \(\dfrac{1}{4}\) < 4b . \(\dfrac{1}{4}\) (nhân cả 2 vế cho \(\dfrac{1}{4}\) )

        a < b

2. 

a. Ta có: a < b 

3a < 3b ( nhân cả 2 vế cho 3)

3a - 7 < 3b - 7 (cộng cả 2 vế cho - 7 )

b. Ta có: a < b

-2a > -2b (nhân cả 2 vế cho -2)

5 - 2a > 5 - 2b ( cộng cẩ 2 vế cho 5)

c. Ta có: a < b 

2a < 2b (nhân cả vế cho 2)

2a + 3 < 2b + 3 (cộng cả 2 vế cho 3)

d. Ta có: a < b

3a < 3b (nhân cả 2 vế cho 3)

3a - 4 < 3b - 4 (cộng cả 2 vế cho -4)

Ta có: 3 < 4

đến đây ko bắt cầu qua đc chắc đề bài sai

A =(a+b-2c) -(-a+b+c) -(2a-b-c)

   = a+b-2c+a-b-c-2a+b+c

   = b-2c

B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)

  = -2a+b-c+b-2c-3a+5a+3c-b

  = b-c

C=(3a-b-2c)-( 2b+3c-a) +(2a-3b)

  = a-b-2c-2b-3c+a+2a-3b

  = -6b-5c

D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)

   = 5a-3b+c+2a-3b+5-b+c-a

   = 6a-7b+2c

\(A=\left(a+b-2c\right)-\left(-a+b+c\right)-\left(2a-b-c\right)\)

\(=a+b-2c+a-b-c-2a+b+c=b-2c\)

\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)

\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)

\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)

\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)

\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)

\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)