\(x-3\frac{1}{3}:\frac{5}{3}=7\)

 \(x-\frac{10}{3}\cdot\frac{3}{5}=7\)

x - 2                      = 7

x                            = 9

\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có

\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)

       \(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
        \(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)

\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)

\(x+\left[x-\frac{3}{2}\right]=0\)

\(\Leftrightarrow x+x-\frac{3}{2}=0\)

\(\Leftrightarrow2x=\frac{3}{2}\Leftrightarrow x=\frac{3}{2}:2=\frac{3}{4}\)

x + (x - \(\frac{3}{2}\)) = 0

<=> x + x - \(\frac{3}{2}\)= 0

<=> 2x                 = \(\frac{3}{2}\)

#hatsunemiku

<=> x                   = \(\frac{3}{4}\)

89,23 : 18 + 90,77 : 18

= (89,23 + 90,77) : 18

= 180 : 18

= 10

\(d,89,23\div18+98,77\div18.\)

\(=89,23\times\frac{1}{18}+99,77\times\frac{1}{18}\)

\(=\frac{1}{18}\left(89,23+90,77\right)\)

\(=\frac{1}{18}.180=\frac{180}{18}=10\)

7000⁴⁴⁴⁴ = ........0

6000⁵⁵⁵⁵ = ........0

mà 5555 > 4444 

=> 7000⁴⁴⁴⁴ < 6000⁵⁵⁵⁵ 

7000⁴⁴⁴⁴ = 1000⁴⁴⁴⁴. 7⁴⁴⁴⁴ 

6000⁵⁵⁵⁵ = 1000⁴⁴⁴⁴.1000¹¹¹¹ . 6⁵⁵⁵⁵ 

=> 7⁴⁴⁴⁴ < 1000¹¹¹¹. 6⁵⁵⁵⁵

=> 7000⁴⁴⁴⁴ < 6000⁵⁵⁵⁵

\(c,\left(23,1\times97,63\right)\div9,763\)

\(=23,1\times10\times9,763\div97,63\)

\(=23,1\times1=23,1\)

\(\frac{3}{2}\cdot\frac{17}{5}+\frac{19}{5}\cdot\frac{3}{2}+\frac{3}{2}\cdot\frac{1}{5}\)

\(=\frac{3}{2}\left[\frac{17}{5}+\frac{19}{5}+\frac{1}{5}\right]\)

\(=\frac{3}{2}\cdot\frac{37}{5}=\frac{111}{10}\)

1) Tính nhanh

\(\frac{3}{2}.\frac{17}{5}+\frac{19}{5}.\frac{3}{2}+\frac{3}{2}.\frac{1}{5}\)

= \(\frac{3}{2}.\left(\frac{17}{5}+\frac{19}{5}+\frac{1}{5}\right)\)

= \(\frac{3}{2}.\frac{37}{5}\)

= \(\frac{111}{10}\)

= \(11,1\)

\(x+9\left(x-\frac{3}{2}\right)=0\)

\(\Rightarrow x+9x-\frac{27}{2}=0\)

\(\Rightarrow10x=\frac{27}{2}\)

\(\Rightarrow x=\frac{27}{2}\div10=\frac{27}{2}.\frac{1}{10}=\frac{27}{20}\)

\(x+9\left[x-\frac{3}{2}\right]=0\)

\(\Leftrightarrow x+9x-\frac{27}{2}=0\)

\(\Leftrightarrow10x=\frac{27}{2}\)

\(\Leftrightarrow x=\frac{27}{2}:10=\frac{27}{2}\cdot\frac{1}{10}=\frac{27}{20}\)

Vậy : ...

Tỉ số giữa số lớn với số bé là 

0,95 = 95/100 = 19/20

Gọi số bé là 19 phần , số lớn là 20 phần

=> Hiệu số phần bằng nhau là :

20 - 19 = 1 (phần)

=> Số lớn là : 

0,95 : 1 x 20 = 19

=> Số bé là :

19 - 0,95 = 18,05 

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