3/5*x+8/21+9=16:7/10

3/5*x+8/21=160/7-9

3/5*x=97/7-8/21

x=283/21:3/5

x=1415/63

\(16\div\left(\frac{3}{5}\times x+8+\frac{9}{21}\right)=\frac{7}{10}\)

\(\Rightarrow\frac{3}{5}x+\frac{59}{7}=\frac{160}{7}\)

\(\Rightarrow\frac{3}{5}x=\frac{101}{7}\)

\(\Rightarrow x=\frac{35}{303}\)

~Study well~

#KSJ

\(16:\left(\frac{3}{5}.x+8+\frac{9}{21}\right)=\frac{7}{10}\)

\(\frac{3}{5}x+\frac{59}{7}=\frac{160}{7}\)

\(\frac{3}{5}x=\frac{101}{7}\)

\(x=\frac{35}{303}\)

5/9x12/7=20/21

6/5:8/3=9/20

9/20x5/12=3/16

15/16:25/24=9/10

14x5/21=10/3

10:5/3=6

5/3:10=1/6

5/9 x 12/7 = 60/63

6/5 : 8/3 = 6/5 x 3/8 = 9/20

9/20 x 5/12 = 3/16

15/16 : 25/24 = 15/16 x 24/25 = 9/10

14 x 5/21 = 60/21

10 : 5/3 = 10 x 3/5 = 6

5/3 : 10 = 5/3 x 1/10 = 1/6

gvsdfgdfgdg

a)    \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)

\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)

\(=\frac{1}{2}+\frac{1}{3}\)

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

b)   \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)

\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)

\(=1\times\frac{1}{2}=\frac{1}{2}\)

c)   \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)

\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)

\(=\frac{247}{420}+\frac{72}{420}\)

\(=\frac{319}{420}\)

còn phần D đâu bn 

tích cho chị nha

 câu a)     = 5/6

 câu b = 1/2

 câu c = 319/420

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

98775 - 32 x 85

=98775 -2720

=96055

 67500 - 24 x 236

= 67500 -5664

=61836

 568 + 101598 : 287

= 568 +354

=922

6875 + 980 -180  

=7855 -180 

=7675

\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)

\(=\dfrac{7}{10}-\dfrac{1}{2}\)

= \(\dfrac{1}{5}\)

\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)

\(=\dfrac{8}{11}+\dfrac{2}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}x\dfrac{8}{5}\)

\(=\dfrac{8}{30}\)

\(=\dfrac{4}{15}\)

\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)

\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)

\(=\dfrac{5}{12}-\dfrac{1}{6}\)

\(=\dfrac{5}{12}-\dfrac{2}{12}\)

\(=\dfrac{3}{12}=\dfrac{1}{4}\)

_{Tín hịu cầu kíu :>>}

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}