\(a,\dfrac{3}{8}=\dfrac{6}{x}\\ \Rightarrow x=6:\dfrac{3}{8}\\ \Rightarrow x=16\\ b,\dfrac{1}{9}=\dfrac{x}{27}\\ \Rightarrow x=\dfrac{1}{9}.27\\ \Rightarrow x=3\\ c,\dfrac{4}{x}=\dfrac{8}{6}\\ \Rightarrow x=4:\dfrac{4}{3}\\ \Rightarrow x=3\\ d,\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Rightarrow3x+6=-4x+20\\ \Rightarrow3x+6+4x-20=0\\ \Rightarrow7x-14=0\\ \Rightarrow7x=14\\ \Rightarrow x=2\)

a: =>6/x=3/8

hay x=16

b: =>x/27=1/9

nên x=3

c: =>4/x=4/3

nên x=3

d: =>3/x-5=-4/x+2

=>3x+2=-4x+20

=>7x=18

hay x=18/7

• \(x\cdot2\frac{4}{9}=4\frac{2}{5}-1\frac{4}{7}\)

\(x\cdot\frac{22}{9}=\frac{22}{5}-\frac{11}{7}\)

\(x\cdot\frac{22}{9}=\frac{99}{35}\)

\(x=\frac{99}{35}:\frac{22}{9}\)

\(x=\frac{81}{70}\)

Vậy \(x=\frac{81}{70}\).

• \(x:10\frac{6}{5}=4\frac{6}{5}+9\frac{7}{5}\)

\(x:\frac{56}{5}=\frac{26}{5}+\frac{52}{5}\)

\(x:\frac{56}{5}=\frac{78}{5}\)

\(x=\frac{78}{5}\cdot\frac{56}{5}\)

\(x=\frac{4368}{25}\)

Vậy \(x=\frac{4368}{25}\).

• \(x:3\frac{10}{3}=9\frac{1}{9}-5\frac{1}{9}\)

\(x:\frac{19}{3}=4\)

\(x=4\cdot\frac{19}{3}\)

\(x=\frac{76}{3}\)

Vậy \(x=\frac{76}{3}\).

(x-1)/3=8/6?????

a, 23/24

b, 23/18

c, 189/8

d, 40/3

\(a,x+\frac{3}{8}=\frac{4}{3}\)

\(x=\frac{4}{3}-\frac{3}{8}\)

\(x=\frac{23}{24}\)

\(b,x-\frac{1}{9}=\frac{7}{6}\)

\(x=\frac{7}{6}+\frac{1}{9}\)

\(x=\frac{23}{18}\)

\(c,x\times\frac{1}{9}=\frac{21}{8}\)

\(x=\frac{21}{8}:\frac{1}{9}\)

\(x=\frac{189}{8}\)

\(d,x:\frac{8}{3}=5\)

\(x=5\times\frac{8}{3}\)

\(x=\frac{40}{3}\)

a: \(P=\frac{x+6\sqrt{x}}{\sqrt{x}-1}\)

\(=\frac{x-\sqrt{x}+7\sqrt{x}-7+7}{\sqrt{x}-1}=\sqrt{x}+7+\frac{7}{\sqrt{x}-1}\)

=>\(P=\sqrt{x}-1+\frac{7}{\sqrt{x}-1}+8\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{7}{\sqrt{x}-1}}+8=2\sqrt7+8\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}-1\right)^2=7\)

=>\(\sqrt{x}-1=\sqrt7\)

=>\(\sqrt{x}=\sqrt7+1\)

=>\(x=\left(\sqrt7+1\right)^2=8+2\sqrt7\) (nhận)

b: \(P=\frac{x-9\sqrt{x}}{\sqrt{x}+3}\)

\(=\frac{x+3\sqrt{x}-12\sqrt{x}-36+36}{\sqrt{x}+3}=\sqrt{x}-12+\frac{36}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{36}{\sqrt{x}+3}-15\ge2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{36}{\sqrt{x}+3}}-15=2\cdot6-15=-3\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}+3\right)^2=36\)

=>\(\sqrt{x}+3=6\)

=>\(\sqrt{x}=3\)

=>x=9