x + 3 và 1/5 = 4
( x - 2/3 ) x 3 = 3/4
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Bài 1
a) 3 2/5 - 1/2
= 17/5 - 1/2
= 34/10 - 5/10
= 29/10
b) 4/5 + 1/5 × 3/4
= 4/5 + 3/20
= 16/20 + 3/20
= 19/20
c) 3 1/2 × 1 1/7
= 7/2 × 8/7
= 4
d) 4 1/6 : 2 1/3
= 25/6 : 7/3
= 25/14
Bài 2
a) 3 × 1/2 + 1/4 × 1/3
= 3/2 + 1/12
= 18/12 + 1/12
= 19/12
b) 1 4/5 - 2/3 : 2 1/3
= 9/5 - 2/3 : 7/3
= 9/5 - 2/7
= 63/35 - 10/35
= 53/35
=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)
\(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)
\(x\) = -\(\dfrac{29}{12}\)
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
A = 2⁵.(-5)² - 8² - 7
= 32.25 - 64 - 7
= 729
= 27²
B = 2³.(-4)² + (-3)².3² - 40
= 8.16 + 9.9 - 40
= 169
= 13²
C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³
= (-5/4)³ . (8/5)³
= (-5/4 . 8/5)³
= (-2)³
D = (-1/4)² : (1/2 - 1/3)
= 1/16 : 1/6
= 3/8
E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³
= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³
= 1/4 + 25 . (15/16)³ : 27/8
= 1/4 + 25 . 3375/4096 : 27/8
= 1/4 + 84375/4096 : 27/8
= 1/4 + 3125/512
= 3253/512
F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8
= 8 + 3.1 - 1 + (4 : 1/2) - 8
= 8 + 3 - 1 + 8 - 8
= 10
a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)
a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
a)\(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\text{⇔}\dfrac{7}{4}x-5=\dfrac{10}{3}\text{⇔}\dfrac{7}{4}x=\dfrac{25}{3}\text{⇔}x=\dfrac{100}{21}\)
b)\(\dfrac{2}{3}x+\dfrac{1}{4}=\dfrac{7}{12}\text{⇔}\dfrac{2}{3}x=\dfrac{1}{3}\text{⇔}x=\dfrac{1}{2}\)
c)\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\text{⇔}\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\text{⇔}x+1=\dfrac{5}{3}\text{⇔}x=\dfrac{2}{3}\)
d)\(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\text{⇔}\dfrac{1}{3}:3x=-\dfrac{21}{4}\text{⇔}\dfrac{1}{9x}=-\dfrac{21}{4}\text{⇔}9x=-\dfrac{4}{21}\text{⇔}x=-\dfrac{4}{189}\)
a, \(1\dfrac{3}{4}x-5=3\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{4}x=5+\dfrac{10}{3}=\dfrac{25}{3}\)
\(\Rightarrow x=\dfrac{25}{3}:\dfrac{7}{4}=\dfrac{100}{21}\)
Vậy ...
b, \(PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{7}{12}-\dfrac{1}{4}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\)
Vậy ....
c, \(PT\Leftrightarrow\dfrac{2}{5}\left(x+1\right)=1-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Rightarrow x+1=\dfrac{2}{3}:\dfrac{2}{5}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{5}{3}-1=\dfrac{2}{3}\)
Vậy ...
d, \(PT\Leftrightarrow\dfrac{1}{3}:3x=-5-\dfrac{1}{4}=\dfrac{1}{9}x=-\dfrac{21}{4}\)
\(\Rightarrow x=-\dfrac{189}{4}\)
Vậy ...
1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
\(x+3\dfrac{1}{5}=4\)
\(x+\dfrac{16}{5}=4\)
\(x=4-\dfrac{16}{5}\)
\(x=\dfrac{4}{5}\)
\(\left(x-\dfrac{2}{3}\right)\times3=\dfrac{3}{4}\)
\(\left(x-\dfrac{2}{3}\right)=\dfrac{3}{4}:3\)
\(x-\dfrac{2}{3}=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}+\dfrac{2}{3}\)
\(x=\dfrac{11}{12}\)
\(x+3\dfrac{1}{5}=4\)
=> \(x+\dfrac{16}{5}=4\)
\(x\) \(=4-\dfrac{16}{5}\)
\(x\) \(=\dfrac{20}{5}-\dfrac{16}{5}\)
\(x\) \(=\dfrac{4}{5}\)
\(\left(x-\dfrac{2}{3}\right) . 3=\dfrac{3}{4}\)
\(\left(x-\dfrac{2}{3}\right)\) \(=\dfrac{3}{4} : 3\)
\(\left(x-\dfrac{2}{3}\right)\) \(=\dfrac{3}{4} : \dfrac{3}{1}\)
\(\left(x-\dfrac{2}{3}\right)\) \(=\dfrac{1}{4}\)
\(x\) \(=\dfrac{1}{4}+\dfrac{2}{3}\)
\(x\) \(=\dfrac{11}{12}\)