Ta có: \(\sqrt{2}>1\)

\(\Rightarrow1+\sqrt{2}>1+1\)

\(\Rightarrow1+\sqrt{2}>2\)

Ta có:\(\sqrt{2}>\sqrt{1}\)

\(\Leftrightarrow1+\sqrt{2}>1+\sqrt{1}=2\)

#)Giải :

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=1\)

\(\Rightarrow a+b+c=a+b-c\Rightarrow c=-c\Rightarrow c-\left(-c\right)=0\Rightarrow c+c=0\Rightarrow c=0\left(đpcm\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)

\(\Rightarrow\frac{a+b+c}{a+b-c}=1\Rightarrow a+b+c=a+b-c\)

\(\Rightarrow a+b+c-a-b+c=0\)

\(\Rightarrow2c=0\Rightarrow c=0\)(đpcm)

\(A^2=\left(2+\sqrt{3}\right)^2=7+\sqrt{48}\)

\(B^2=\left(\sqrt{2}+3\right)^2=11+\sqrt{72}\)

\(\hept{\begin{cases}7< 11\\\sqrt{48}< \sqrt{72}\end{cases}\Leftrightarrow}7+\sqrt{48}< 11+\sqrt{72}\)

\(\Rightarrow A< B\)

Ta có:\(2+\sqrt{3}< 2+\sqrt{4}=4=\sqrt{1}+3< \sqrt{2}+3\)

   \(\Rightarrow2+\sqrt{3}< \sqrt{2}+3\)

hoa anh dao la sakura

Ta có:\(\left|x\right|+0,573=2\)

\(\Leftrightarrow\left|x\right|=1,427\)

Vì x là số dương nên x > 0

Vậy \(x=1,427\)

 Tính tổng

S = 1 + (-2) + 3 + (-4) + ... + 99 + (-100)

S = 1 - 2 + 3 - 4 + ... + 99 - 100

S = ( 1-2 ) + ( 3 - 4 ) + ....  + ( 99 - 100 )

S = -1 - 1 - 1 - .... - 1  ( 50 số -1 )

S = -50

Study well 

\(S=1+\left(-2\right)+3+\left(-4\right)+...+99+\left(-100\right)\)

\(S=1-2+3-4+...+99-100\)

\(S=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)

\(S=-1-1-1-...-1\)(50 số -1)

\(S=-50\)

a,  0,4 : x = x : 0,9

<=> x² = 0,4 . 0,9

<=> x² = 0,36

<=> x = 0,6 hoặc -0,6

b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow10=26\div\left(2x-1\right)\)

\(\Leftrightarrow2x-1=\frac{13}{5}\)

\(\Leftrightarrow2x=\frac{18}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\)

c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow6x+7=4\)

\(\Leftrightarrow6x=-3\)

\(\Leftrightarrow x=\frac{-1}{2}\)

d, \(\frac{37-x}{x+13}=\frac{3}{7}\) 

\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)

\(\Leftrightarrow259-7x=3x+39\)

\(\Leftrightarrow-10x=-220\)

\(\Leftrightarrow x=22\)

1.Gọi \(\frac{a}{b}=\frac{c}{d}=k\) 

\(\Rightarrow a=bk\)

      \(c=dk\)  

Ta có

\(\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b.\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)

\(\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d.\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\frac{a}{a-b}=\frac{c}{c-d}\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\)

\(\frac{a}{c}=\frac{bk}{dk}=\frac{b}{d}\left(1\right)\)

\(\frac{a-b}{c-d}=\frac{bk-b}{dk-d}=\frac{b.\left(k-1\right)}{d.\left(k-1\right)}=\frac{b}{d}\left(2\right)\)

Từ ( 1 ) và ( 2  ) \(\Rightarrow\frac{a}{c}=\frac{a-b}{a-c}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

Các phần khác em cũng đặt = k  và làm tương tự nha bây giờ ah đang vội nên không thể làm cho e đc sorry

Study well 

\(E=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}\)

\(=\frac{1}{2}.\frac{3}{4}\)

\(=\frac{3}{8}\)