1.khang dinh A,B,D dung

2,x-(1-x)=5+(-1+x)

  x-1+x=5-1+x

  2x-1=4+x

2x-x=4+1

x=5

Vay x=5

co bai kho hoi mik nhe

uk.Mơn bạn

Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)

Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$

Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:

$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$

Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$

Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$

$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$

Nên $Q \leq  \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)

Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$

Bài 1: 

a: \(\dfrac{x}{25}=\dfrac{-4}{-x}=\dfrac{4}{x}\)

\(\Leftrightarrow x^2=100\)

=>x=10 hoặc x=-10

b: \(\dfrac{x}{10}=\dfrac{11}{x+1}\)

\(\Leftrightarrow x^2+x-110=0\)

=>(x+11)(x-10)=0

=>x=10 hoặc x=-11

8)

a)6-|x|=2

=>6-x=2

x=6-2

x=4

b)6+|x|=2

=>6+x=2

x=2-6

x=-4

a) |x| + |-5| = |-37|

<=> |x| + 5 = 37

<=> |x| = 37 - 5 = 32

=> x \(\in\) {32 ; -32}

b)|-6| . |x| = |54|

<=> 6 . |x| = 54

|x| = 54 : 6 = 9

=> x \(\in\){9;-9}

c) |x| > 21

Có |x| \(\ge\) 0 > 21

=> |x| \(\in\) { 22 ; 23 ; 24 ; 25 ; ....}

=> x \(\in\) { 22; -22 ; 23; -23; 24; -24; 25; -25; ....}

a (a+b+c) + (a-b+c) - (b+c-a) - (a-b-c)

 =a+b+c+a-b+c-b-c+a-a+b+c

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

=     a x 2 + 0 + c x 2

=       2 x (a + c)

\(\left(a+b-c\right)+\left(a-b+c\right)-\left(a-b-c\right)\)

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

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

\(=a+b+c\)

\(x-\left(+5\right)-\left[\left(x+11\right)-\left(x-11\right)\right]\)

\(=x-5-\left[x+11-x+11\right]\)

\(=x-5-\left[x-x+11+11\right]\)

\(=x-5-22\)

\(=x-27\)

\(a-\left\{\left(a-5\right)-\left[\left(a+9\right)-\left(-a+1\right)\right]\right\}\)

\(=a-\left\{a-5-\left[a+9+a-1\right]\right\}\)

\(=a-\left\{a-5-\left[a+a+9-1\right]\right\}\)

\(=a-\left\{a-5-\left[2a+8\right]\right\}\)

\(=a-\left\{a-5-2a-8\right\}\)

\(=a-a+5+2a+8\)

\(=2a+13=2\left(a+6\right)+1\)

a: A={-4;-3;-2;-1;0;1;2;3;4}

b: B={10;11;12;...;18;19;20}

toán 12 khó quá chưa học tới