Tìm x:(2x/7-5):(-8)=0,70
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 5(2x -1) - 4(8 - 3x) = 7
<=> 10x - 5 - 32 + 12x = 7
<=> 22x = 44
<=> x =2
Vậy x = 2 là nghiệm phương trình
b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4)
<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4
<=> -11x - 39 = - 6
<=> -11x = 33
<=> x = -3
Vậy x = -3 là nghiệm phương trình
\(a,10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
\(b,14x-35-35x+10+10x-14=x-2-x-4\)
\(-11x-39=-6\)
\(-11x=-33\)
\(x=3\)
`(x-5)-(2x+7)=-8`
`=> x-5-2x-7=-8`
`=> x-5-2x-7+8=0`
`=> -x -4=0`
`=> -x=4`
`=> x=-4`
Vậy `x=-4`
a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)
\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)
\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)
\(\dfrac{x}{3}=\dfrac{29}{15}\)
\(x=5,8\)
b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)
\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{7}{2}\)
\(x=\dfrac{7}{4}\)
a)\(60x^2+35x-60x^2+15x=100\)
35x+15x=100
50x=100 =>x=2
b)\(10x^2-35x+16x-10x^2=5\)
-35x+16x=5
-19x=5 =>x=-5/19
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
\(\left(x-5\right)-\left(2x+7\right)=-8\)
\(\Rightarrow x-5-2x-7=-8\)
\(-x\) \(=-8+5+7\)
\(-x\) \(=4\)
\(\Rightarrow-x\) \(=-4\)
Vậy x= -4
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
\(\left(\dfrac{2x}{7}-5\right):\left(-8\right)=0,7\)
\(\Rightarrow\dfrac{2x}{7}-5=0,7.\left(-8\right)\)
\(\Rightarrow\dfrac{2x}{7}=-5,6+5\)
\(\Rightarrow\dfrac{2x}{7}=-0,6\Rightarrow x=-0,6.\dfrac{7}{2}=-2,1\)
( \(\dfrac{2x}{7}\) - 5) : (-8) = 0,70
\(\dfrac{2x}{7}\) - 5 = 0,7 .(-8)
\(\dfrac{2x}{7}\) - 5 = -5,6
\(\dfrac{2x}{7}\) = -5, 6 + 5
\(\dfrac{2}{7}x\) = -,06
\(x\) = -0,6 : \(\dfrac{2}{7}\)
\(x\) = -2,1