tìm x
a, (x - 2/7 ) ( x + 3/4 )=0
b,2/3x -2/5=1/2x -1/3
c,1/3x +2/5 (x+1 )=0
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\(\dfrac{1}{6}\) - 0,4 . \(\dfrac{5}{8}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\)
= \(\dfrac{2}{12}\) - \(\dfrac{3}{12}\) + \(\dfrac{6}{12}\)
= \(\dfrac{5}{12}\)
Đề vẫn giải được nhưng số khá xấu, không phù hợp với lớp 7. Bạn xem lại xem đã đúng đề chưa vậy>
\(\dfrac{-19}{2}\) < a < \(\dfrac{-20}{3}\)
- \(\dfrac{57}{6}\) < a < \(\dfrac{-40}{6}\)
a = - 56/6; -55/6; -54/6; - 53/6; - 52/6; - 51/6; ......; -42/6; -41/6
b, \(\dfrac{1}{3}\) < \(\dfrac{4}{b}\) < \(\dfrac{1}{2}\)
\(\dfrac{4}{12}\) < \(\dfrac{4}{b}\)< \(\dfrac{4}{8}\)
b = 11; 10; 9
2(x-3)+3(x+1)=4x-1
=>2x-6+3x+3=4x-1
=> 2x+3x-4x=-1+6-3
=> x(2+3-4)=2
=>x=2
Vậy x=2
\(2\times\left(x-3\right)+3\times\left(x+1\right)=4x-1\)
\(2x-6+3x+3=4x-1\)
\(\Rightarrow2x-6+3x+3-\left(4x-1\right)=0\)
\(2x-6+3x+3-4x+1=0\)
\(2x+3x-4x-6+3+1=0\)
\(\left(2+3-4\right)x-\left(6-3-1\right)=0\)
\(x-2=0\)
\(x=0+2\)
\(x=2\)
2(x-3)+3(x+1)=4x-1
=>2x-6+3x+3=4x-1
=> 2x+3x-4x=-1+6-3
=> x(2+3-4)=2
=>x=2
Vậy x=2
a, (x- \(\dfrac{2}{7}\))(x+\(\dfrac{3}{4}\))
⇔ \(\left[{}\begin{matrix}x-\dfrac{2}{7}=0\\x+\dfrac{3}{4}=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
b, \(\dfrac{2}{3}\) x - \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\) x - \(\dfrac{1}{3}\)
⇔ \(\dfrac{2}{3}\) x - \(\dfrac{1}{2}\) x = \(\dfrac{2}{5}\) - \(\dfrac{1}{3}\)
⇔ \(\dfrac{x}{6}\) = \(\dfrac{1}{15}\)
⇔ x = \(\dfrac{1}{15}\) x 6
x = \(\dfrac{2}{5}\)
c, \(\dfrac{1}{3}\) x + \(\dfrac{2}{5}\) (x+1)= 0
⇔ \(\dfrac{5x+6x+6}{15}\) = 0
⇔ 11x + 6 = 0
⇔ x = - \(\dfrac{6}{11}\)
tìm x
a, (x - 2/7 ) ( x + 3/4 )=0
<=> x - 2/7= 0 hoặc x+3/4 = 0
=> X = 2/7 hoặc x = -3/4
c,1/3x +2/5 (x+1 )=0
=> 1/3x + 2/5x + 2/5.1 = 0
=> x.(1/3 + 2/5 ) +2/5 = 0
=> x.11/15 + 2/5 = 0
=> x.11/15 = 0 - 2/5
=> x.11/15 = -2/5
=> x = -2/5 : 11/15
=> x = -6/11
Vậy x = -6/11