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

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

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

b) \(\frac{a}{a+c}=\frac{b}{b+d}\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\Rightarrow\frac{bk}{b}=\frac{bk+dk}{1.b+1.d}\Rightarrow k=k\left(đpcm\right)\)

Vậy \(\frac{a}{a+c}=\frac{b}{b+d}\)

a, \(\sqrt{0,01}-\sqrt{0,05}=\sqrt{\frac{1}{100}}-\sqrt{\frac{5}{100}}\)

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

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

\(b,0,5.\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-\frac{\sqrt{1}}{\sqrt{4}}\)

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

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

Giải

\(\sqrt{0,01}-\sqrt{0,05}=\frac{1-\sqrt{5}}{10}\)

\(0.5\times\sqrt{100}-\sqrt{\frac{1}{4}}=\frac{29}{2}\)

Lớp 7 chưa hok cộng trừ  đâu nhá

\(A=\left|\frac{1}{3}x+4\right|+\frac{2}{3}\)

\(\left|\frac{1}{3}x+4\right|\ge0\Rightarrow\left|\frac{1}{3}x+4\right|+\frac{2}{3}\ge\frac{2}{3}\)

\(\Rightarrow A\ge\frac{2}{3}\)

dấu "=" xảy ra khi : 

|1/3x + 4| = 0

=> 1/3x + 4 = 0

=> 1/3x = -4

=> x = -12

\(B=\left|x-6\right|+\left|x+\frac{5}{4}\right|\)

\(\left|x-6\right|\ge6-x\)

\(\left|x+\frac{5}{4}\right|\ge x+\frac{5}{4}\)

\(\Rightarrow\left|x-6\right|+\left|x+\frac{5}{4}\right|\ge6-x+x+\frac{5}{4}\)

\(\Rightarrow B\ge\frac{29}{4}\)

dấu "=" xảy ra khi : 

x - 6 < 0 => x < 6

x + 5/4 > 0 => x > -5/4 

vậy -5/4 < x < 6

a) |5x - 1| - x = 2x + 3

<=> |5x - 1| = 2x + 3 + x

<=> |5x - 1| = 3x + 3

<=> 5x - 1 = 3x + 3 hoặc 5x - 1 = -(3x + 3)

       5x - 1 - 3x = 3            5x - 1 + 3x = -3

       2x - 1 = 3                   8x - 1 = -3

       2x = 3 + 1                  8x = -3 + 1

       2x = 4                        8x = -2

       x = 2                           x = -2/8 = -1/4

=> x = 2 hoặc x = -1/4

b) Ta có: |2x + 1| \(\ge\)0 \(\forall\)x

        |x - 3| \(\ge\)0 \(\forall\)x

     |2x+ 3| \(\ge\)0  \(\forall\)x

=> |2x + 1| + |x - 3| + |2x + 3| \(\ge\)0 \(\forall\)x

=> x - 5 \(\ge\)0 \(\forall\)x => x \(\ge\)5 \(\forall\)x

Với x \(\ge\)5 

=> 2x + 1 + x - 3 + 2x + 3 = x - 5

=> 4x + 1 = x - 5

=> 4x - x = -5 - 1

=> 3x = -6

=> x = -2 (ktm)

Vậy ko có giá trị thõa mãn

\(a,1,\left(27\right)=1+\frac{27}{99}=1+\frac{3}{11}=\frac{14}{11}\)

\(0,\left(423\right)=\frac{423}{999}=\frac{47}{111}\)

\(2,\left(15\right)=2+\frac{15}{99}=2+\frac{5}{33}=\frac{71}{33}\)

\(b,3,04\left(31\right)=3+\frac{431-4}{9990}=3+\frac{427}{9990}=\frac{3397}{9990}\)

\(1,8\left(34\right)=1+\frac{834-8}{990}=1+\frac{413}{495}=\frac{908}{495}\)

A lớn hơn B

\(a,x-\frac{2}{3}=\frac{3}{5}-\frac{1}{4}\)

\(\Rightarrow x-\frac{2}{3}=\frac{7}{20}\)

\(\Rightarrow x=\frac{61}{60}\)

\(b,-\frac{2}{3}.x+\frac{1}{5}=\frac{1}{10}\)

\(\Rightarrow-\frac{2}{3}.x=-\frac{1}{10}\)

\(\Rightarrow x=\frac{3}{20}\)

\(c,\frac{4}{5}-\frac{1}{2}.x=\frac{1}{10}\)

\(\Rightarrow\frac{1}{2}.x=\frac{7}{10}\)

\(\Rightarrow x=\frac{7}{5}\)

\(d,2-\left(x+1\right)=1\)

\(\Rightarrow x+1=1\)

\(\Rightarrow x=0\)

\(e,\frac{1}{2}:x-\frac{5}{4}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}:x=\frac{7}{4}\)

\(\Rightarrow x=\frac{2}{7}\)

\(f,4:\left(1+x\right)=\frac{2}{4}+\frac{3}{2}\)

\(\Rightarrow4:\left(1+x\right)=2\)

\(\Rightarrow1+x=2\)

\(\Rightarrow x=1\)

a)x-2/3=7/20

x=7/20+2/3

x=61/60

b)-2/3.x=1/10-1/5

x=-1/10:-2/3

x=3/20

c)1/2.x=4/5-1/10

x=7/10:1/2

x=7/5

d)x+1=2-1

x=1-1

x=0

e)1/2:x=1/2+5/4

x=7/4.1/2

x=7/8

f)4:(1+x)=2

1+x=2.4

x=8-1

x=7