\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

nhớ làm giúp mình nhá mai mình phải đi học r :<

=> 1/3x-1/4+x^2-9/16=0

=(1/3x+x/2)+....

các bước sau tự giải

Ta có: \(\hept{\begin{cases}\left(\frac{1}{3}x-\frac{1}{4}\right)^2\ge0;\forall x\\\left(x^2-\frac{9}{16}\right)^4\ge0;\forall x\end{cases}}\)\(\Rightarrow\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4\ge0;\forall x\)

Do đó \(\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{3}x-\frac{1}{4}\right)^2=0\\\left(x^2-\frac{9}{16}\right)^4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\pm\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow x=\frac{3}{4}\)

Vậy \(x=\frac{3}{4}\)

\(\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{3}x-\frac{1}{4}=0\\x^2-\frac{9}{16}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{3}x=\frac{1}{4}\\x^2=\frac{9}{16}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\end{cases}\Rightarrow}x=\frac{3}{4}}\)\(\Rightarrow\hept{\begin{cases}\frac{1}{3}x-\frac{1}{4}=0\\x^2-\frac{9}{16}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{3}x=\frac{1}{4}\\x^2=\frac{9}{16}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\x=\pm\frac{3}{4}\end{cases}\Rightarrow}x=\frac{3}{4}}\)

Vậy x=3/4

hok tốt

a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(x-3\right)\left(x^2+x\cdot3+3^2\right)\)

\(=x^3-3^3=x^3-27\)

b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x-2\right)\left(x^2+x\cdot2+2^2\right)\)

\(=x^3-2^3=x^3-8\)

c) Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)\)

\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)

\(=x^3+4^3=x^3+64\)

d) Ta có: \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=\left(x-3y\right)\left[x^2+x\cdot3y+\left(3y\right)^2\right]\)

\(=x^3-\left(3y\right)^3=x^3-27y^3\)

e) Ta có: \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\)

\(=\left(x^2-\frac{1}{3}\right)\left[\left(x^2\right)^2+x^2\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3\)

\(=x^6-\frac{1}{27}\)

f) Ta có: \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)

\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x\cdot2y+\left(2y\right)^2\right]\)

\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\frac{1}{27}x^3+8y^3\)

\(\left(\left(\frac{3}{4}\right)^3\right)^2=\left(\frac{9}{16}\right)^x\)

=>\(\left(\frac{3}{4}\right)^6=\left(\left(\frac{3}{4}\right)^2\right)^x\)

=>\(\left(\frac{3}{4}\right)^6=\left(\frac{3}{4}\right)^{2x}\)

=>6=2x

=>x=3

Sorry, cho mình làm lại:

\(\left(\left(\frac{3}{4}\right)^3\right)^2=\left(\frac{16}{9}\right)^x\)

=>\(\left(\frac{3}{4}\right)^6=\left(\left(\frac{4}{3}\right)^2\right)^x\)

=>\(\left(\frac{3}{4}\right)^6=\left(\frac{4}{3}\right)^{2x}\)

=>\(\left(\frac{3}{4}\right)^6=\frac{1}{\left(\frac{3}{4}\right)^{2x}}\)

=>\(\left(\frac{3}{4}\right)^6=\left(\frac{3}{4}\right)^{-2x}\)

=>6=-2x

=>-6=2x

=>x=-3

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)

Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)

=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)

=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)

=> \(x^2-4x-2x+8-x-2=-2x\)

=> \(x^2-5x+6=0\)

=> \(\left(x-2\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)

=> x = 3 .

Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)

b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)

Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)

=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)

=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)

=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)

=> \(x\left(x+12\right)=192\)

=> \(x^2+12x-192=0\)

=> \(x^2+2x.6+36-228=0\)

=> \(\left(x+6\right)^2=288\)

=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )

Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a, |x-7|=|9-234|

=> |x-7|=|-225|

=> |x-7|=225

=>\(\orbr{\begin{cases}x-7=225\\x-7=-225\end{cases}}\)=>\(\orbr{\begin{cases}x=232\\x=-218\end{cases}}\)

b, \(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

=>\(\left(\frac{x}{8}\right)^2=\left(\frac{1}{2}\right)^2\)

=>\(\orbr{\begin{cases}\frac{x}{8}=\frac{1}{2}\\\frac{x}{8}=\frac{-1}{2}\end{cases}}\)=>\(\orbr{\begin{cases}x.2=8\\x.2=-8\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

c, (8-3²)²+(x-39)²=10

=>(-1)²+(x-39)²=10

=>(x-39)²=10-1

=>(x-39)²=9

=>(x-39)²=3²

=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)

a)\(\left|x-7\right|=\left|9-234\right|\)

\(\Rightarrow\left|x-7\right|=\left|-225\right|\)

\(\Rightarrow\left|x-7\right|=225\)

\(\Rightarrow x-7=\pm225\)

Vậy \(x\in\left\{232;-218\right\}\)

b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)

\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)

Vậy \(x\in\left\{\pm4\right\}\)

c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow1+\left(x-39\right)^2=10\)

\(\Rightarrow\left(x-39\right)^2=9\)

\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow x-39=\pm3\)

Vậy \(x\in\left\{42;36\right\}\)