a) \(\left(\frac{1}{5}+\frac{3}{5}\right)\times\frac{1}{4}\)

\(=\frac{4}{5}\times\frac{1}{4}\)

\(=\frac{4}{20}=\frac{1}{5}\)

\(\left(\frac{1}{5}+\frac{3}{5}\right)\times\frac{1}{4}=\frac{1}{5}\times\frac{1}{4}+\frac{3}{5}\times\frac{1}{4}\)

                                  \(=\frac{1}{20}+\frac{3}{20}\)

                                    \(=\frac{4}{20}=\frac{1}{5}\)

b) \(\left(\frac{7}{9}-\frac{2}{9}\right)\times\frac{3}{2}\)

\(=\frac{5}{9}\times\frac{3}{2}\)

\(=\frac{15}{18}=\frac{5}{6}\)

\(\left(\frac{7}{9}-\frac{2}{9}\right)\times\frac{3}{2}=\frac{7}{9}\times\frac{3}{2}-\frac{2}{9}\times\frac{3}{2}\)

                                  \(=\frac{21}{18}-\frac{6}{18}\)

                                   \(=\frac{15}{18}=\frac{5}{6}\)

c) \(\left(\frac{11}{13}-\frac{9}{13}\right)\times\frac{39}{21}\)

\(=\frac{2}{13}\times\frac{39}{21}\)

\(=\frac{78}{273}=\frac{2}{7}\)

\(\left(\frac{11}{13}-\frac{9}{13}\right)\times\frac{39}{21}=\frac{11}{13}\times\frac{39}{21}-\frac{9}{13}\times\frac{39}{21}\)

                                       \(=\frac{429}{273}-\frac{351}{273}\)

                                         \(=\frac{78}{273}=\frac{2}{7}\)

a,\(\left(\frac{1}{5}+\frac{3}{5}\right)x\frac{1}{4}\)

=\(\frac{4}{5}x\frac{1}{4}\)

=\(\frac{4}{20}\)

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

,\(\left(\frac{1}{5}+\frac{3}{5}\right)x\frac{1}{4}\)

=\(\frac{4}{5}x\frac{1}{4}\)

=\(\frac{4x1}{5x4}\)

=\(\frac{1x1}{5x1}\)

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

\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)

\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)

\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)

\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)

dấu chấm là gì vậy

dấu chấm là dấu x trong cấp trên

xin loi nhe minh chi biet lam bai c thoi a hihihihihihhi

c) (2008 - 8) : 2 + 1 = 1001

4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)

\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)

\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)

\(\Leftrightarrow-22x-60=-75x-30\)

\(\Leftrightarrow-22x+75x=-30+60\)

\(\Leftrightarrow53x=30\)

\(\Leftrightarrow x=\dfrac{30}{53}\)

Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)

5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)

\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)

\(\Leftrightarrow10x-6-21x+3=60\)

\(\Leftrightarrow-11x-3=60\)

\(\Leftrightarrow-11x=63\)

\(\Leftrightarrow x=-\dfrac{63}{11}\)

Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)

`9,x^3+x^2-2=0`

`x^3-x^2+2x^2-2=0`

`<=>x^2(x-1)+2(x-1)(x+1)=0`

`<=>(x-1)(x^2+2x+2)=0`

`<=>x=1`

`14,x^2-2x+1=0`

`<=>(x-1)^2=0`

`<=>x-1=0`

`<=>x=1`

`15,x^3+3x^2+3x+1=0`

`<=>(x+1)^3=0`

`<=>x+1=0`

`<=>x=-1`

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}=\dfrac{11}{11}\times\dfrac{3}{7}=1\times\dfrac{3}{7}=\dfrac{3}{7}\)

\(\dfrac{3}{5}\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{3}{5}\times\dfrac{5}{9}=\dfrac{1}{3}\)

`( 6 / 11 + 5 / 11 ) xx 3 / 7 = 11 / 11 xx 3 / 7 = 1 xx 3 / 7 = 3 / 7`

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`3 / 5 xx 7 / 9 - 3 / 5 xx 2 / 9 = 3 / 5 xx ( 7 / 9 - 2 / 9 ) = 3 / 5 xx 5 / 9 = 1 / 3`

(6/11+5/11)x3/7

=11/11x3/7

=33/77

3/5x7/9-3/5x2/9

=21/63-6/63

=15/63

Bài 6: 

1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)

\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)

\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)

\(\Leftrightarrow-14x-9=0\)

\(\Leftrightarrow-14x=9\)

\(\Leftrightarrow x=-\dfrac{9}{14}\)

Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)

`1)2x(x-5)-(x+3)^2=3x-x(5-x)`

`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`

`<=>x^2-16x-9=x^2-2x`

`<=>14x=-9`

`<=>x=-9/14`

Lời giải:
a.

$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=[9^2-2.18]^2-2.18^2=1377$

Nếu $x\geq y$ thì:

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$

$=\sqrt{9^2-4.18}(9^2-18)=189$

Nếu $x< y$ thì $x^3-y^3=-189$

b.

$A=(x+y)^2-6(x+y)+y-5$

$=(-9)^2-6(-9)+y-5=130+y$

Chưa đủ cơ sở để tính biểu thức.

a) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=9^3-3\cdot18\cdot9=243\)

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

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

\(=\left(9^2-2\cdot18\right)^2-2\cdot18^2\)

\(=45^2-2\cdot324\)

=1377

10/3.x+47/4=-53/4

10/3.x=-53/4-47/4

10/3.x=-25

x=-25:10/3

x=-15/2

\(3\dfrac{1}{3}.x+11\dfrac{3}{4}=\left(-13,25\right)\) 

  \(\dfrac{10}{3}.x+\dfrac{47}{4}=\dfrac{-53}{4}\) 

           \(\dfrac{10}{3}.x=\dfrac{-53}{4}-\dfrac{47}{4}\) 

           \(\dfrac{10}{3}.x=-25\) 

                  \(x=-25:\dfrac{10}{3}\) 

                  \(x=\dfrac{-15}{2}\)