a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

\(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\dfrac{9}{4}\)

\(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(\dfrac{1}{3}-\dfrac{3}{2}.x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{3}{2}\) hoặc \(\dfrac{1}{3}-\dfrac{3}{2}x=\dfrac{-3}{2}\)

\(\dfrac{3}{2}x=\dfrac{3}{2}+\dfrac{1}{3}\) hoặc \(\dfrac{3}{2}x=\dfrac{-3}{2}+\dfrac{1}{3}\)

\(\dfrac{3}{2}x=\dfrac{9}{6}+\dfrac{2}{6}\) hoặc \(\dfrac{3}{2}x=-\dfrac{9}{6}+\dfrac{2}{6}\)

\(\dfrac{3}{2}x=\dfrac{11}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{-7}{6}\)

\(x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}\) hoặc \(x=\dfrac{-7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}\)

\(x=\dfrac{11}{9}\) hoặc \(x=-\dfrac{7}{9}\)

Vậy...

Mình sửa lại ( từ dòng 4 trở xuống ):

... \(\dfrac{3}{2}x=\dfrac{1}{3}-\dfrac{3}{2}\) hoặc \(\dfrac{3}{2}x=\dfrac{1}{3}+\dfrac{3}{2}\)

\(\dfrac{3}{2}x=\dfrac{2}{6}-\dfrac{9}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{2}{6}+\dfrac{9}{6}\)

\(\dfrac{3}{2}x=-\dfrac{7}{6}\) hoặc \(\dfrac{3}{2}x=\dfrac{11}{6}\)

\(x=-\dfrac{7}{6}:\dfrac{3}{2}=\dfrac{-7}{6}.\dfrac{2}{3}\) hoặc \(x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}.\dfrac{2}{3}\)

\(x=\dfrac{-7}{9}\) hoặc \(x=\dfrac{11}{9}\)

Vậy...

a) \(x+1\dfrac{4}{7}=2\dfrac{2}{3}\)

\(x+\dfrac{11}{7}=\dfrac{8}{3}\)

\(x=\dfrac{8}{3}-\dfrac{11}{7}\)

\(x=\dfrac{23}{21}\)

b) \(3\dfrac{9}{12}:x=4\dfrac{1}{6}\)

\(\dfrac{15}{4}:x=\dfrac{25}{6}\)

\(x=\dfrac{15}{4}:\dfrac{25}{6}\)

\(x=\dfrac{9}{10}\)

$x-3)2=9$

⇒[2x−3=32x−3=−3⇒[x=3x=0⇒[2x−3=32x−3=−3⇒[x=3x=0

Vậy x = 3 hoặc x = 0

\(\left(2x-3\right)^2=9\)

\(\left(2x-3\right)^2=3^2\)

⇒\(2x-3=+-3\)

\(TH1:2x-3=3\text{⇒}x=3\)

\(TH2:2x-3=-3\text{⇒}x=0\)

ta có: \(\left(2x-3\right)^2\)=\(3^2\)

2x-3=3

2x=3+3

2x=6

x=6:2

x=3

vậy x==3

a,=0,9:(4/5.1,25+1:1/3)
=0,9:(1+3)

=0,9:4

=0,225

b,=9,6:6-0,6

=1,6-0,6

=1

c,=7/2.11/4-7/2.5/4

=7/2.(11/4-5/4)

=7/2.3/2

=21/4

a; Xét ΔBAC có MN//BC

nên AM/AB=AN/AC

=>AM/20=15/20

=>AM=15

b: Xét ΔABC có MN//BC

nên AN/NC=AM/MB

=>AN/NC=3/2

=>AN/3=NC/2

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

\(\dfrac{AN}{3}=\dfrac{NC}{2}=\dfrac{AN+NC}{3+2}=\dfrac{5}{5}=1\)

Do đó: NC=2

c: Xét ΔBCA có MN//BC

nên MN/BC=AM/AB

=>MN/6=8/12=2/3

hay MN=4

a) Thay x = 81 vào A ta có:

\(A=\dfrac{4\sqrt{81}}{\sqrt{81}-5}=\dfrac{4\cdot9}{9-5}=\dfrac{4\cdot9}{4}=9\)

b) \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{x+\sqrt{x}-2}\left(x\ne1;x\ge0\right)\)

\(B-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

c) \(\dfrac{A}{B}< 4\) khi

\(\dfrac{4\sqrt{x}}{\sqrt{x}-5}:\dfrac{\sqrt{x}}{\sqrt{x}+2}< 4\)

\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-5}< 4\)

\(\Leftrightarrow\dfrac{4\sqrt{x}+8-4\left(\sqrt{x}-4\right)}{\sqrt{x}-5}< 0\)

\(\Leftrightarrow\dfrac{24}{\sqrt{x}-5}< 0\)

\(\Leftrightarrow\sqrt{x}-5< 0\)

\(\Leftrightarrow x< 25\)

Kết hợp với đk: 

\(0\le x< 5\)