b) 16:( 3/5* x + 8 /21 +9) = 7/10
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\(16:\left(\frac{3}{5}x+\frac{8}{21}+9\right)=\frac{7}{10}\)
\(\frac{3}{5}x+\frac{8}{21}+9=\frac{160}{7}\)
\(\frac{3}{5}x=\frac{283}{21}\)
\(x=\frac{1415}{63}\)
Vậy ...
\(16\div\left(\frac{3}{5}\times x+8+\frac{9}{21}\right)=\frac{7}{10}\)
\(\Rightarrow\frac{3}{5}x+\frac{59}{7}=\frac{160}{7}\)
\(\Rightarrow\frac{3}{5}x=\frac{101}{7}\)
\(\Rightarrow x=\frac{35}{303}\)
~Study well~
#KSJ
\(16:\left(\frac{3}{5}.x+8+\frac{9}{21}\right)=\frac{7}{10}\)
\(\frac{3}{5}x+\frac{59}{7}=\frac{160}{7}\)
\(\frac{3}{5}x=\frac{101}{7}\)
\(x=\frac{35}{303}\)
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) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
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}
b, 3/5 + 4/7 + 2/8 + 10/25 + 9/21 + 28/16
= 3/5 + 4/7 + 2/8 + 2/5 + 3/7 + 14/8
= (3/5 + 2/5) + ( 4/7 + 3/7) + ( 2/8 + 14/8)
= 1 + 1 + 7/4
= 2 + 7/4 = 15/4
c , 8/7 + 7/6 + 5/8 + 10/12 + 24/28 + 6/16
= c , 8/7 + 7/6 + 5/8 + 5/6 + 6/7 + 1/2
= (8/7 + 6/7) + (7/6 + 5/6) + 5/8 + 1/2
= 14/7 + 12/6 + 5/8 + 1/2
= 2 + 2 + 5/8 + 1/2
= 4 + 9/8 = 41/8
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
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