Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3x+2 - 3x =24
..............
➙ x =1
vậy x=1
3x+2- 3x= 24
(=)3x*32-3x=24
(=)3x*32-3x*1=24
3x*(32-1)=24
3x*(9-1)=24
3x*8=24
3x=24/8=3
=)x=1
=)
a) x^2 +3x-2x-6=0
x^2 + x = 6
x^2 + 0.5x + 0.5x = 6
x (x + 0.5) + 0.5 (x + 0.5) =5.75
(x+0.5)^2 = 5.75
`1/8+(2x+5)/24=(3x+5)/2+x/6`
`=>3+2x+5=12(3x+5)+4x`
`=>2x+8=36x+60+4x`
`=>2x+8=40x+60`
`=>38x=-52`
`=>x=-26/19`
Vậy `x=-26/19`
\(3x-\dfrac{1}{2}=x+\dfrac{1}{2}\)
\(\Rightarrow3x-x=\dfrac{1}{2}+\dfrac{1}{2}\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(4-\dfrac{7}{2}x-1=5\)
\(\Rightarrow3-\dfrac{7}{2}x=5\)
\(\Rightarrow\dfrac{7}{2}x=-2\)
\(\Rightarrow x=\dfrac{7}{2}:-2\)
\(\Rightarrow x=-\dfrac{7}{4}\)
\(A=\left(\frac{1+2x}{2.\left(2+x\right)}-\frac{x}{3.\left(x-2\right)}+\frac{2x^2}{3.\left(4-x^2\right)}\right).\frac{24-12x}{6+13x}\)
\(=\left[\frac{3.\left(1+2x\right)\left(2-x\right)-2x\left(x+2\right)+4x^2}{2.3.\left(x+2\right)\left(2-x\right)}\right].\frac{24-12x}{6+13x}\)
\(=\frac{6+9x-6x^2-2x^2-4x+4x^2}{6.\left(4-x^2\right)}.\frac{24-12x}{6+13x}\)
\(=\frac{6+5x-4x^2}{6.\left(4-x^2\right)}.\frac{12.\left(2-x\right)}{6+13x}\) \(=\frac{\left(6+5x-4x^2\right).2}{\left(x+2\right)\left(6+13x\right)}=\frac{12+10x-8x^2}{13x^2+32x+12}\)
\(\Leftrightarrow\)3x. 32 - 3x = 24
\(\Leftrightarrow\)3x. ( 32 - 1 ) = 24
\(\Leftrightarrow\)3x . ( 9 - 1 ) = 24
\(\Rightarrow\)3x . 8 = 24
3x = 24 : 8
3x = 31
\(\Rightarrow\)x = 1