\(C=\left|-3\left(\dfrac{-13}{15}-\dfrac{17}{21}\right)\right|-\left|\dfrac{-13}{15}+\dfrac{17}{7}\right|+\left(-12+\dfrac{35}{3}\right):\left|-\dfrac{7}{6}\right|\\ =\left|-3.-\dfrac{176}{105}\right|-\left|-\dfrac{6}{35}\right|+\left(-\dfrac{1}{3}\right):\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{1}{3}:\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{2}{7}\\ =\dfrac{170}{35}-\dfrac{2}{7}=\dfrac{32}{7}.\)

tính giúp mình với mình đang cần gấp

cảm ơn.

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{17}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{13}\)

\(\dfrac{9}{17}\times\dfrac{21}{13}+\dfrac{9}{17}\times\dfrac{5}{13}-\dfrac{9}{17}\times2\)

\(=\dfrac{9}{17}\times\left(\dfrac{21}{13}+\dfrac{5}{13}-2\right)\)

\(=\dfrac{9}{17}\times\left(\dfrac{26}{13}-2\right)=\dfrac{9}{17}\times\left(2-2\right)\)

\(=\dfrac{9}{17}\times0=0\)

= 9/17 x ( 21/13 + 5/13 - 2 )

= 9/17 x 0

= 0

=9/17 x (21/13 + 5/13 - 2)

=9/17 x (26/13 - 2)

=9/17 x (13 - 2)

=9/17 x 11

= 99/17

Tick cho mik nhé!!!

a) Áp dụng t/x dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-5}=\dfrac{3x}{6}=\dfrac{2z}{-10}=\dfrac{3x-2z}{6+10}=\dfrac{48}{16}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.3=9\\z=3.\left(-5\right)=-15\end{matrix}\right.\)

b) \(\dfrac{x}{10}=\dfrac{y}{-13}=\dfrac{z}{17}=\dfrac{2y}{-26}=\dfrac{3z}{51}=\dfrac{2y-3z}{-26-51}=\dfrac{77}{-77}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.\left(-1\right)=-10\\y=\left(-13\right).\left(-1\right)=13\\z=17.\left(-1\right)=-17\end{matrix}\right.\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-5}\Rightarrow\dfrac{3x}{6}=\dfrac{y}{3}=\dfrac{2z}{-10}\)

Áp dụng t/c của DTSBN, ta có: \(\dfrac{3x-2z}{6-\left(-10\right)}=\dfrac{48}{16}=3\)

\(\dfrac{x}{2}=3\Rightarrow x=6\)

\(\dfrac{y}{3}=3\Rightarrow y=9\)

\(\dfrac{z}{-5}=3\Rightarrow z=-15\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{-13}=\dfrac{z}{17}=\dfrac{2y-3z}{2\cdot\left(-13\right)-3\cdot17}=\dfrac{77}{-77}=-1\)

Do đó: x=-10; y=13; z=-17

\(=\dfrac{-3}{17}-\dfrac{2}{13}+\dfrac{20}{17}-\dfrac{11}{13}\)

=1-1

=0

\(-\left(\dfrac{3}{17}+\dfrac{2}{13}\right)-\left(\dfrac{-20}{17}+\dfrac{11}{13}\right)\)

\(=-\dfrac{3}{17}-\dfrac{2}{13}-\dfrac{20}{17}-\dfrac{11}{13}\)

\(=-\left(\dfrac{3}{17}+\dfrac{20}{17}\right)-\left(\dfrac{2}{13}+\dfrac{11}{13}\right)\)

\(=-\dfrac{23}{17}-1\)

\(=-\dfrac{40}{17}\)

\(-\left(\dfrac{3}{17}+\dfrac{2}{13}\right)-\left(\dfrac{-20}{17}+\dfrac{11}{13}\right)\)

⇔ \(\dfrac{-3}{17}-\dfrac{2}{13}+\dfrac{20}{17}-\dfrac{11}{13}\)

⇔ \(\left(\dfrac{-3}{17}+\dfrac{20}{17}\right)+\left(\dfrac{-2}{13}-\dfrac{11}{13}\right)\)

⇔ \(1+\left(-1\right)\)

⇔ \(0\)

   \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)

= \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)

= \(\dfrac{-4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)

= \(\dfrac{-4}{13}.\dfrac{17}{17}\)

= \(\dfrac{-4}{13}.1\)

= \(\dfrac{-4}{13}\)

= \(\dfrac{-4.5-12.4}{13.17}\)

=\(\dfrac{-4\left(5+12\right)}{13.17}\)

=\(\dfrac{-4.17}{13.17}\)

=\(\dfrac{-4}{13}\)

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)

biểu thức đâu

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

khó quá

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30