(1/1-1/3+1/3-1/5+...+1/19-1/20)*63-x=20

(1/1-1/20)*63-x=20

19/20*63-x=20

1177/20-x=20

58,85-x=20

x=58,85-20

x=38,85

kich dung nha

              Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

              \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

             \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

              \(2A=1-\frac{1}{21}\)

              \(A=\frac{20}{21}:2\)

            \(A=\frac{10}{21}\)

          => \(\frac{10}{21}\)x 63 - x = 20

         => 30 - x = 20

        => x = 30 - 20 = 10

      Vậy x = 10

     Ủng hộ mk nha !!! ^_^

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=1-\frac{1}{21}\)

\(A=\frac{20}{21}:2\)

> $\frac{10}{21}$1021 x 63 - x = 20

         => 30 - x = 20

        => x = 30 - 20 = 10

      Vậy x = 10

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}

\(\frac{-4}{7}-x=\frac{3}{5}-2x\)

\(\frac{-4}{7}-\frac{-3}{5}=-2x+x\)

                 \(\frac{1}{35}=x.\left(-2+1\right)\)

   \(\frac{1}{35}=-1.x\)

       \(x=\frac{1}{35}:-1\)

       \(x=\frac{-1}{35}\)

Học tốt nha!!!

(1/2*X+2/1/4)*-2/3=2/5/6

(1/2*X+9/4)*-2/3=17/6

(1/2*X+9/4)=-17/4

1/2*X=-13/2

X=-13

mik cũng ra -13

Tham khảo ở đây nhé bn: olm.vn/hoi-dap/question/686545.html, mk lm r`

\(\text{ta có: }-x.x=-9.4=-36\text{ nên: }x^2=36\text{ hay }x=6\text{ hoặc -6}\)

vì cộng đến 20 mà tổng bằng 20 nên:

x+(x+1)+(x+2)+...+19=0 do đó: x+19=0 hay: x=-19

cảm ơn nha :33

Lời giải :

Theo đề bài ta có \(\frac{x}{\frac{5}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{6}{5}}\Leftrightarrow\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}\)

Đặt \(\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}=k\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{5k}{2}\\z=\frac{6k}{5}\end{cases}}\)

Mặt khác : \(\frac{x}{2}=\frac{z-28}{3}\)

\(\Leftrightarrow3x-2z=-56\)

\(\Leftrightarrow3\cdot\frac{5k}{2}-2\cdot\frac{6k}{5}=-56\)

\(\Leftrightarrow k=\frac{-560}{51}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1400}{51}\\y=\frac{-2240}{153}\\z=\frac{-224}{17}\end{cases}}\)

\(B=x+y-z=\frac{-1400}{51}+\frac{-2240}{153}-\frac{-224}{17}=\frac{-4424}{153}\)

Ta có :

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\frac{9}{12}-\frac{10}{12}\le\frac{x}{12}< 1-\left(\frac{8}{12}-\frac{3}{12}\right)\)

\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)

\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\frac{x}{12}\in\left\{-\frac{1}{12};\frac{0}{12};\frac{1}{12};\frac{2}{12};\frac{3}{12};\frac{4}{12};\frac{5}{12};\frac{6}{12}\right\}\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5;6\right\}\)

Ủng hộ mk nha !!! ^_^