B1:Ta có: A:4 dư 3=>A-3 chia hết cho 4=> A+1 chia hết cho 4.

           A:5 dư 4=>A-4 chia hết cho 5=> A+1 chia hết cho 5.

=>A+1 chia hết cho 4,5.

mà (4,5)=1

=>A+1 chia hết cho 4.5=20
=>A+1 chia hết cho 20

=>A+1=20k(k thuộc Z)

=>A=20k-1

Vậy A=20k-1(k thuộc Z)

l-i-k-e cho mình đi rồi mình làm tiếp B2 cho.

\(S=\frac{a}{1+b}+\frac{b}{1+a}+\frac{1}{a+b}=\frac{a^2}{a+ab}+\frac{b^2}{b+ab}+\frac{1}{a+b}\)

\(S\ge\frac{\left(a+b\right)^2}{a+b+2ab}+\frac{1}{a+b}\ge\frac{\left(a+b\right)^2}{a+b+\frac{\left(a+b\right)^2}{2}}+\frac{1}{a+b}\)

\(S\ge\frac{2\left(a+b\right)}{a+b+2}+\frac{1}{a+b}=2-\frac{4}{a+b+2}+\frac{1}{a+b}\)

Đặt \(a+b=t\Rightarrow0< t\le1\)

\(S\ge\frac{5}{3}+\frac{t+3}{3t}-\frac{4}{t+2}=\frac{5}{3}+\frac{t^2-7t+6}{3t\left(t+2\right)}=\frac{5}{3}+\frac{\left(6-t\right)\left(1-t\right)}{3t\left(t+2\right)}\ge\frac{5}{3}\)

\(S_{min}=\frac{5}{3}\) khi \(t=1\Leftrightarrow x=y=\frac{1}{2}\)

a, x= -9/2

b, x= 2

kb nha

a) \(\left|x+5\right|-x+23=27\)

\(\Leftrightarrow\left|x+5\right|-x+23-27=0\)

\(\Leftrightarrow\left|x+5\right|-x+\left(23-27\right)=0\)

\(\Leftrightarrow\left|x+5\right|-x+\left(-4\right)=0\)

\(\Leftrightarrow\left|x+5\right|-x-4=0\)

\(\Leftrightarrow\left|x+5\right|-\left(x+4\right)=0\)

\(\Leftrightarrow\left|x+5\right|=x+4\)

Xét trường hợp 1:

\(x+5=x+4\)

\(\Leftrightarrow x+5-\left(x+4\right)=0\)

\(\Leftrightarrow x+5-x-4=0\)

\(\Leftrightarrow1=0\left(loai\right)\)

Xét trường hợp 2:

\(x+5=-\left(x+4\right)\)

\(\Leftrightarrow x+5-\left[-\left(x+4\right)\right]=0\)

\(\Leftrightarrow x+5+x+4=0\)

\(\Leftrightarrow2x+9=0\)

\(\Leftrightarrow2x=-9\)

\(\Leftrightarrow x=\dfrac{-9}{2}\left(thoa\right)\)

Vậy \(x=\dfrac{-9}{2}\)

Với x;y dương, ta có BĐT:

\(x^5+y^5\ge x^2y^2\left(x+y\right)\)

Thật vậy, BĐT tương đương:

\(x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow x^4\left(x-y\right)-y^4\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (luôn đúng)

Áp dụng:

\(\Rightarrow A\le\dfrac{ab}{a^2b^2\left(a+b\right)+ab}+\dfrac{bc}{b^2c^2\left(b+c\right)+bc}+\dfrac{ca}{c^2a^2\left(c+a\right)+ca}\)

\(A\le\dfrac{1}{ab\left(a+b\right)+1}+\dfrac{1}{bc\left(b+c\right)+1}+\dfrac{1}{ca\left(c+a\right)+1}\)

\(A\le\dfrac{abc}{ab\left(a+b\right)+abc}+\dfrac{abc}{bc\left(b+c\right)+abc}+\dfrac{abc}{ca\left(c+a\right)+abc}=\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=1\)

\(1,\\ a,X=\left\{3;4\right\};\left\{2;3;4\right\};\left\{1;2;3;4\right\}\\ b,X=\left\{2;4\right\}\\ X=\left\{2\right\}\\ X=\left\{4\right\}\\ X=\varnothing\)

\(2,\\ a,A=\left\{-3;-2;0;1;2;3;4\right\}\\ B=\left\{0;1;2;3;4;6;9;10\right\}\\ b,A=\left\{1;2;3;4;5\right\}\\ B=\left\{1;2;3;6;9\right\}\)

a,123-5(x+4)=28

(=)5(x+4)=85

(=) x+4=17

(=) x=13

b,10+2x=2.\(\left(3^2-1\right)\)

(=) 10+2x=16

(=) 2x = 6 

(=)x=3

A, 123-5.(x+4)=38

            5.(x+4)=123-38

           5.(x+4) =85

               (x+4)=85:5

               (x+4)=17

                x      =17-4

                x      =13

B, 10+2x=2(\(3^2-1\))

     10+2x=2.8

     10+2x=16

           2x=16-10

           2x=6

             x=6:2

            x  =3

a, 123 - 5(x + 4) = 38

=>5(x + 4) = 85

=> x + 4 = 17

=> x = 13

vậy_

10 + 2x = 2(3² - 1)

=> 10 + 2x = 2.8

=> 10 + 2x = 16

=> 2x = 6

=> x = 3

vậy_

TC: a/b=b/c

=>