-37.(63 + 12) + 63.(37 - 12)

= -37.63 + (-37).12 + 63.37 - 63.12

= [-37.63 + 63.37] + [-37.12 - 63.12]

= 0 + [-37.12 - 63.12]

= -37.12 - 63.12

= (-37 - 63).12

= -100.12

= -1200

Chọn `A.\sqrt{36}+\sqrt{63}>10`
Vì `\sqrt{63}>\sqrt{16}=4`
`=>\sqrt{36}+\sqrt{63}>6+4=10`

Chọn A

A

Cần CM: \(\frac{1}{9-a}-\frac{12}{a^2+63}\ge\frac{1}{144}a^2-\frac{1}{16}\) (1) 

\(\Leftrightarrow\)\(\frac{a^2+12a-45}{\left(9-a\right)\left(a^2+63\right)}\ge\frac{1}{144}a^2-\frac{1}{16}\)

\(\Leftrightarrow\)\(144\left(a^2+12a-45\right)\ge\left(a-3\right)\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\)

\(\Leftrightarrow\)\(\left(a-3\right)\left[144\left(a+15\right)-\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\right]\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)\left(a^4-6a^3+36a^2-234a+459\right)\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)^2\left(a^3-3a^2+27a+153\right)\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)^2\left[\left(a-3\right)^2\left(a+3\right)+36a+126\right]\ge0\) ( đúng )

Do đó (1) đúng => \(\Sigma_{cyc}\frac{1}{9-a}-\Sigma_{cyc}\frac{12}{a^2+63}\ge\frac{1}{144}\left(a^2+b^2+c^2\right)-\frac{3}{16}=0\)

\(\Rightarrow\)\(\Sigma_{cyc}\frac{12}{a^2+63}\le\Sigma_{cyc}\frac{1}{9-a}\le\Sigma_{cyc}\frac{1}{a+b}\) ( do \(a+b+c\le9\) ) 

Dấu "=" xảy ra khi a=b=c=3 

Ta có: \(\frac{1}{a+b}+\frac{1}{b+c}\ge2\sqrt{\frac{1}{a+b}\frac{1}{b+c}}=2\frac{1}{\sqrt{\left(a+b\right)\left(b+c\right)}}\ge\frac{4}{a+2b+c}\)

Tương tự có: \(\frac{1}{b+c}+\frac{1}{a+c}\ge\frac{4}{a+2c+b}\)

\(\frac{1}{a+b}+\frac{1}{a+c}\ge\frac{4}{b+2a+c}\)

\(\Rightarrow\frac{1}{a+b}+\frac{1}{c+b}+\frac{1}{a+c}\ge2\left(\frac{1}{b+2a+c}+\frac{1}{a+2b+c}+\frac{1}{b+2c+a}\right)\)

Ta CM: \(\frac{1}{b+2a+c}\ge\frac{6}{a^2+63}\). Thật vậy:

\(\frac{1}{b+2a+c}\ge\frac{6}{a^2+63}\)\(\Leftrightarrow a^2+63\ge6b+12a+6c\)\(\Leftrightarrow2a^2+b^2+c^2+36-6b-12a-6c\ge0\)

\(\Leftrightarrow2\left(a-3\right)^2+\left(b-3\right)^2+\left(c-3\right)^2\ge0\) ( luôn đúng)

Dấu '=' xảy ra <=> a=b=c=3

Vậy \(\frac{1}{b+2a+c}+\frac{1}{a+2b+c}+\frac{1}{b+2c+a}\ge\frac{6}{a^2+63}+\frac{6}{b^2+63}+\frac{6}{c^2+63}\)

=> đpcm

3:

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

=a-b+c-c+b+a

=2a

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

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

\(=b-a\)

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

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

\(=-a+b-c\)

2:

a: \(12-\left(-12\right)=12+12=24\)

b: \(-\left(35-49\right)+\left(27-49\right)=-35+49+27-49\)

=-35+27

=-8

c: \(47-\left(59-63\right)+\left(63-47\right)\)

\(=47-59+63+63-47\)

=126-59

=67

d: \(-\left(-20\right)-\left(-30\right)-70\)

=20+30-70

=50-70

=-20

290

= 290

12+20+63+95+100= 290

\(\frac{-63}{108}\)= \(\frac{-7}{12}\)

\(\frac{-33}{-77}\)= \(\frac{3}{7}\)

\(\frac{-5}{10}\)=\(\frac{-1}{2}\)

\(\frac{14}{63}\)=\(\frac{2}{9}\)

\(\frac{-15}{25}\)=\(\frac{-3}{5}\)

\(\frac{-45}{18}\)=\(\frac{-5}{2}\)

\(\frac{12}{15}\)=\(\frac{4}{5}\)

\(\frac{20}{25}\)=\(\frac{4}{5}\)

\(\frac{31}{12}\):Là phân số tối giản

t.i.c.k nha                                           

ta có:x+63=135

         x=135-63=72

vậy x=72

x+63=123+12

 x+63= 135

x        =  135-63

  x      =   72

ủng hộ nhé

72 nha ban!

là 519

Ta có : 24 x 18 + 12 x 63 + 12

= 12 x 36 + 12 x 63 + 12 x 1

= 12 x ( 36 + 63 + 1) 

= 12 x 100

= 1200

Ta có : 24 x 18 + 12 x 63 + 12

= 12 x 36 + 12 x 63 + 12 x 1

= 12 x ﴾ 36 + 63 + 1﴿

= 12 x 100

= 1200