x × 0,75 + 3/4 × 5 + 2 : 4/3 = 10,5
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a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)
b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{x\left(x-2\right)\left(x+2\right)}{-5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)
c: \(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}\)
d: \(\left(x^2-3x+2\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
=(x-2)(x^2-1)
=>\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)
1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
a) \(0,75\times26+49\times\frac{3}{4}+\frac{3}{4}\times24+\) \(0,75\)
\(=\) \(0,75\times\left(26+1\right)+\frac{3}{4}\times\left(49+24\right)\)
\(=\) \(\frac{3}{4}\times27+\frac{3}{4}\times73\)
\(=\) \(\frac{3}{4}\times\left(27+73\right)\)
\(=\)\(\frac{3}{4}\times100=75\)
b) câu b,c làm tương tự
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
\(a,-\dfrac{3}{5}-x=-0,75\\ -\dfrac{3}{5}-x=-\dfrac{3}{4}\\ x=-\dfrac{3}{5}-\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{3}{20}\\ ---\\ b,1\dfrac{4}{5}=-0,15-x\\ \dfrac{9}{5}=-\dfrac{3}{20}-x\\ x=-\dfrac{3}{20}-\dfrac{9}{5}\\ x=-\dfrac{3}{20}-\dfrac{36}{20}\\ x=-\dfrac{39}{20}\\ ----\\ c,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}+\dfrac{4}{7}\\ \dfrac{33}{10}-x=\dfrac{26}{21}\\ x=\dfrac{33}{10}-\dfrac{26}{21}\\ x=\dfrac{433}{210}\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
a. \(\frac{5}{12}:x=\left(\frac{5}{6}-\frac{1}{2}\right)\times\frac{5}{4}\)
\(=>\)\(\frac{5}{12}:x=\frac{1}{3}\times\frac{5}{4}=\frac{5}{12}\)
\(=>x=\frac{5}{12}:\frac{5}{12}=1\)
b.\(x:0,15=\left(2,8+0,38\right):0,75\)
\(=>x:0,15=3,18:0,75=4,24\)
\(=>x=4,24\times0,15=0,636\)
a) Ta có: 36-x=44
nên x=36-44=-8
Vậy: x=-8
b) Ta có: \(\dfrac{5}{6}-x=\dfrac{7}{4}-0.75-3\)
\(\Leftrightarrow\dfrac{5}{6}-x=\dfrac{7}{4}-\dfrac{3}{4}-3=1-3=-2\)
hay \(x=\dfrac{5}{6}+2=\dfrac{17}{6}\)
a) 36-x=44 b) 5/6-x=7/4-0,75-3
=> x= 36-44 => 5/6-x= -2
=> x= -8 => x= 5/6-(-2)
=> x=17/6
\(x.0,75+\frac{3}{4}.5+2:\frac{4}{3}=10,5\)
= \(x.\frac{3}{4}+\frac{15}{4}+2.\frac{3}{4}=\frac{42}{4}\)
= \(\frac{3}{4}x+\frac{15}{4}+\frac{6}{4}=\frac{42}{4}\)
= \(\frac{3}{4}x=\frac{42}{4}-\frac{15}{4}-\frac{6}{4}=\frac{21}{4}\)
= x = \(\frac{21}{4}:\frac{3}{4}=\frac{21}{3}=7\)
Xx0,75+3/4x5+2:4/3=10,5
x=1,5x5+2:4/3
x=7,5+2:4/3
x=9,5:4/3
x=7,125