042557

1/9 + x = 2/18 + 3/4

x=(2/18+3/4)-1/9

x=3/4

________

x ÷ 1/2 = 3/4

x=3/4x1/2

x=3/8

x = 3/8

Mọi người tk mình đi mình đang bị âm nè!!!!!!

Ai tk mình mình tk lại nha !!!

a.1/4

b.-7/144

c.25/21

d.17/7

e.80/63

f.9/7

a)1/4

b)-7/144

c)-25/21

d)17/7

e)16/63

f)9/7

Tự làm đi, lười thế

a) x= 1/4

b) x= -7/144

c) x= 25/6

d) x=17/7

e) x= 80/63

f) x= -45/14

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

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

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

\(a.\)

\(\frac{4}{9}.x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{9}:\frac{4}{9}\)

\(\Rightarrow x=\frac{1}{9}.\frac{9}{4}\)

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

Vậy \(x=\frac{1}{4}\)

\(b.\)

\(\frac{7}{18}.x-\frac{2}{3}=\frac{5}{18}\)

\(\Rightarrow\frac{7}{18}.x=\frac{5}{18}+\frac{2}{3}\)

\(\Rightarrow\frac{7}{18}.x=\frac{5}{18}+\frac{12}{18}\)

\(\Rightarrow\frac{7}{18}.x=\frac{17}{18}\)

\(\Rightarrow x=\frac{17}{18}:\frac{7}{18}\)

\(\Rightarrow x=\frac{17}{18}.\frac{18}{7}\)

\(\Rightarrow x=\frac{17}{7}\)

Vậy \(x=\frac{17}{7}\)

\(c.\)

\(x:-\frac{1}{12}=\frac{7}{12}\)

\(\Rightarrow x=\frac{7}{12}.-\frac{1}{12}\)

\(\Rightarrow x=-\frac{7}{144}\)

Vậy \(x=-\frac{7}{144}\)

\(d.\)

\(\frac{4}{9}-\frac{7}{8}.x=-\frac{2}{3}\)

\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}-\left(-\frac{2}{3}\right)\)

\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}+\frac{2}{3}\)

\(\Rightarrow\frac{7}{8}.x=\frac{4}{9}+\frac{6}{9}\)

\(\Rightarrow\frac{7}{8}.x=\frac{10}{9}\)

\(\Rightarrow x=\frac{10}{9}:\frac{7}{8}\)

\(\Rightarrow x=\frac{10}{9}.\frac{8}{7}\)

\(\Rightarrow x=\frac{80}{63}\)

Vây \(x=\frac{80}{63}\)

\(e.\)

\(-\frac{5}{14}:x=-\frac{3}{10}\)

\(\Rightarrow x=-\frac{5}{14}:-\frac{3}{10}\)

\(\Rightarrow x=-\frac{5}{14}.-\frac{10}{3}\)

\(\Rightarrow x=\frac{5}{7}.\frac{5}{3}\)

\(\Rightarrow x=\frac{25}{21}\)

Vậy \(x=\frac{25}{21}\)

\(g.\)

\(\frac{1}{6}+\frac{-5}{7}:x=-\frac{7}{18}\)

\(\Rightarrow-\frac{5}{7}:x=-\frac{7}{18}-\frac{1}{6}\)

\(\Rightarrow-\frac{5}{7}:x=-\frac{7}{18}-\frac{3}{18}\)

\(\Rightarrow-\frac{5}{7}:x=-\frac{10}{18}\)

\(\Rightarrow-\frac{5}{7}:x=-\frac{5}{9}\)

\(\Rightarrow x=-\frac{5}{7}:-\frac{5}{9}\)

\(\Rightarrow x=-\frac{5}{7}.-\frac{9}{5}\)

\(\Rightarrow x=\frac{9}{7}\)

Vậy \(x=\frac{9}{7}\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)