\(Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{6\left(x^2+xy+yz+zx\right)}+\sqrt{6\left(y^2+xy+yz+zx\right)}+\sqrt{z^2+xy+yz+zx}}\)

\(\Leftrightarrow Q=\frac{3x+3y+2z}{\sqrt{3\left(x+y\right).2\left(x+z\right)}+\sqrt{3\left(y+x\right).2\left(y+z\right)}+\sqrt{\left(z+x\right).\left(z+y\right)}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(y+x\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}}\)

\(\Rightarrow Q\ge\frac{3x+3y+2z}{\frac{9x+9y+6z}{2}}=\frac{2}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)và  \(z=2\)

\(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=5-\frac{7}{6}\)

\(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=\frac{23}{6}\)

\(x+\frac{4}{3}=\frac{23}{6}:\frac{7}{4}=\frac{46}{21}\)

\(x=\frac{46}{21}-\frac{4}{3}\)

\(x=\frac{6}{7}\)

Vậy ....................................

có ngay em ơi!

Số cần tìm : 

77,4 : 2,58 x 1,5 = 45

Đáp số :  ....

\(\frac{x}{2}=\frac{y}{7}\Leftrightarrow y=\frac{7}{2}x\)

\(xy=x.\frac{7}{2}x=\frac{7}{2}x^2=1400\Leftrightarrow x^2=400\)

\(\Leftrightarrow\orbr{\begin{cases}x=20\Rightarrow y=70\\x=-20\Rightarrow y=-70\end{cases}}\).

ta có x.y=1400

=>y=\(\frac{1400}{x}\)(1)

ta có :\(\frac{x}{2}=\frac{y}{7}\)

=>7x=2y(2)

thay (1) vào (2), ta được:

7x=2.\(\frac{1400}{x}\)

=>7x-\(\frac{2800}{x}\)=0

=>\(\frac{7x^2-2800}{x}=0\)(x\(\ne0\))

=>7x²-2800=0

=>7x²=2800

=>x²=400

=>x=\(\pm20\)

với x=20 =>y=\(\frac{1400}{20}=70\)

với x=-20=>y=\(\frac{1400}{-20}=-70\)

vậy ...

Trả lời:

Đặt \(\frac{x}{2}=\frac{y}{7}=k\left(k\ne0\right)\)(*)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=7k\end{cases}}\)

Ta có:  x . y = 1400

=> 2k . 7k = 1400

=> 14k² = 1400

=> k² = 100

=> k = 10 hoặc k = - 10

Thay k = 10 vào (*), ta được:

\(\Rightarrow\hept{\begin{cases}x=20\\y=70\end{cases}}\)

Thay k = - 10 vào (*), ta được:

\(\Rightarrow\hept{\begin{cases}x=-20\\y=-70\end{cases}}\)

\(x^2+4x+y^2-2xy+x^2+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x+2\right)^2=0\)

vì \(\left(x-y\right)^2\ge0;\left(x+2\right)^2\ge0\)nên

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-y=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=-2\end{cases}\Rightarrow}x=y=-2}\)

a, 2015 - 2 × X  = 101

2 x X = 2015 - 101 

2 x X = 1914

      X = 1914 : 2 

      X = 957

1a. 2015 - 2 x X = 101

                 2 x X = 2015 - 101

                       X = 1914 : 2

                       X = 957

b. 7 - (11 + X - 13) : \(2\frac{2}{3}\)= 2

(11 + X - 13)  : \(2\frac{2}{3}\)         =  7 - 2

11 + X - 13                            = 5 x \(2\frac{2}{3}\)

11 + X                                   = \(\frac{40}{3}\)+ 13

X                                           = \(\frac{79}{3}\)- 11

X                                            = \(\frac{46}{3}\)

c. 2012 - 5 x X = 17

              5 x X   = 2012 - 17

                   X    = 1995 : 5

                   X    = 399

d. \(5\frac{3}{4}\)+ (X - 2) x \(\frac{1}{3}\)= \(17\frac{3}{4}\) 

(X - 2) x \(\frac{1}{3}\)                    = \(17\frac{3}{4}\)- \(5\frac{3}{4}\)

X - 2                                    = 12 : \(\frac{1}{3}\)

X                                          = 36 + 2

X                                          = 38

e. 12 x (X - 5) + 73 = 20

12 x (X - 5)             = 20 - 73

X - 5                       = -53 : 12

X                            = \(\frac{-53}{12}\)+ 5 

X                            = \(\frac{7}{12}\)

a )

2015 - 2 x X = 101

2 x X = 2015 - 101

2 x X = 1914

X = 1914 : 2

X = 957

b )

7 - ( 11 + X - 13 ) : \(2\frac{2}{3}\) = 2

( 11 + X - 13 ) : \(2\frac{2}{3}\) = 7 - 2

( 11 + X - 13 ) : \(2\frac{2}{3}\) = 5

11 + X - 13 = 5 x \(2\frac{2}{3}\)

11 + X - 13 = \(13\frac{1}{3}\)

X = \(13\frac{1}{3}\) + 13 - 11

X = \(15\frac{1}{3}\)

c )

2012 - 5 x X = 17

5 x X = 2012 - 17

5 x X = 1995

X = 1995 : 5

X = 399

d ) 

\(5\frac{3}{4}\) + ( X - 2 ) x \(\frac{1}{3}\) = \(17\frac{3}{4}\)

( X - 2 ) x \(\frac{1}{3}\) = \(17\frac{3}{4}+5\frac{3}{4}\)

( X - 2 ) x \(\frac{1}{3}\) = \(23\frac{1}{2}\)

X - 2 = \(23\frac{1}{2}\div\frac{1}{3}\)

X - 2 = 47

X = 47 + 2

X = 49

e ) Mình không làm đực câu này đâu

Học tốt nhé bạn !

\(2\frac{3}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)

\(=\frac{11}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)

\(=\frac{11}{8}-\frac{3}{14}+\frac{1}{3}\)

\(=\frac{251}{168}\)

Bài 1 : a, thực hiện phép tính : 

\(2\frac{3}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)

\(=\frac{11}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)

\(=\frac{1}{2}×\left(\frac{11}{4}-\frac{3}{7}\right)+\frac{1}{3}\)

\(=\frac{1}{2}×\frac{65}{28}+\frac{1}{3}\)

\(=\frac{65}{56}+\frac{1}{3}\)

\(=\frac{251}{168}\)

b , Tìm x biết : 

a, 435- ( x + 16 ) = 425 : 17 

435 - ( x + 16 ) = 25 

x + 16 = 435 - 25 

x + 16 = 410 

x  = 410 - 16 

x = 394 

Vậy x = 394 

b, ( x + 3/4 ) × 7/4 = 5 - 7/6 

( x + 3/4 ) × 7/4 = 23/6 

x + 3/4 = 23/6 : 7/4 

x + 3/4 = 23/6 × 4/7 

x + 3/4 = 46/21 

x = 46/21 - 3/4 

x = 121/84

Vậy x = 121/84 

Theo đề bài

\(\frac{a}{5}=\frac{b}{3}=\frac{c}{2}\Rightarrow\frac{a}{5}.\frac{b}{3}=\left(\frac{c}{2}\right)^2\Rightarrow\frac{a.b}{15}=\frac{c^2}{4}=\frac{a.b-c^2}{15-4}=\frac{11}{11}=1\)

\(\Rightarrow\frac{c^2}{4}=1\Rightarrow c^2=4\Rightarrow c=\pm2\)

+ Với c=-2

\(\Rightarrow\frac{a}{5}=\frac{b}{3}=\frac{-2}{2}=-1\Rightarrow a=-5;b=-3\)

+ Với c=2

\(\Rightarrow\frac{a}{5}=\frac{b}{3}=\frac{2}{2}=1\Rightarrow a=5;b=3\)