1. Tìm xeQ :
a, 2/3 x = -4/27
b, -1 3/5 x = 1 1/15
giúp e với ạ câu b là hỗn số ạ
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a)\(|a|=\frac{1}{5}\)
\(\Rightarrow\hept{\begin{cases}a=\frac{1}{5}\\a=-\frac{1}{5}\end{cases}}\)
Vậy \(a\in\left\{\frac{1}{5};-\frac{1}{5}\right\}\)
\(a,30+X=120:5+27\\ 30+X=24+27\\ 30+X=51\\ X=51-30=21\\ ---\\ b,40-3\times X=13\\ 3\times X=40-13=27\\ X=\dfrac{27}{3}=9\\ ---\\ 2\times X-8=16\\ 2\times X=16+8\\ 2\times X=24\\ X=\dfrac{24}{2}=12\\ \\---\\ \dfrac{1}{2}\times X-\dfrac{1}{3}=\dfrac{1}{4}\\ \dfrac{1}{2}\times X=\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{7}{12}\\ X=\dfrac{7}{12}:\dfrac{1}{2}=\dfrac{7}{6}\)
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
\(\text{Phần a :}\)
\(\frac{2}{3}.x=\frac{-4}{27}\)
\(\Rightarrow x=\frac{-4}{27}:\frac{2}{3}=\frac{-2}{9}\)
\(\text{Phần b :}\)
\(-1\frac{1}{3}.x=1\frac{1}{15}\)
\(\Rightarrow\frac{-4}{3}.x=\frac{16}{15}\)
\(\Rightarrow x=\frac{16}{15}:\frac{-4}{3}=\frac{-4}{5}\)
a, \(\left|2x+1\right|=3\)
=> 2x + 1 = 3 hoặc 2x + 1 = -3
=> 2x = 3 - 1 hoặc 2x = -3 - 1
=> 2x = 2 hoặc 2x = -4
=> x = 1 hoặc x = -2
b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)
=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)
=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)
=> \(\frac{5}{2}x=\frac{16-9}{6}\)
=> \(\frac{5}{2}x=\frac{7}{6}\)
=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)
c, \(3\cdot5^{x-3}+1=16\)
=> 3 . 5x-3 = 16 - 1
=> 3 . 5x-3 = 15
=> 5x-3 = 15 : 3
=> 5x-3 = 5
=> x - 3 = 5 : 5
=> x - 3 = 1
=> x = 1 + 3 = 4
d, \((x-1)^2=25\)
=> \((x-1)^2=5^2\)
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)
=> 4 + |3x + 1| = -196
=> |3x + 1| = -196 - 4 = -200
=> |3x + 1| = -200
Không thỏa mãn điều kiện
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
a)\(\frac{2}{3}x=-\frac{4}{27}\)
\(\Leftrightarrow x=-\frac{4}{27}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{2}{9}\)
b)\(-1\frac{3}{5}x=1\frac{1}{15}\)
\(\Leftrightarrow-\frac{2}{5}x=\frac{16}{15}\)
\(\Leftrightarrow x=-\frac{8}{3}\)
#H
ac=11