Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
a. \(\frac{5x-2}{3}=\frac{5x-3x}{2}\)
\(\Leftrightarrow2.\left(5x-2\right)=3.\left(5x-3x\right)
\)
\(\Leftrightarrow10x-4=15x-9x\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy...
b. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\left(1\right)\)
MC = 36.
pt (1) <=>
\(\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
=> 3.(10x+3) = 36 + 4(6+8x)
<=> 30x+9 = 36+24+32x
<=> -2x = 51
<=> x = \(\frac{-51}{2}\)
Vậy...
c. \(\frac{7x-1}{6}+2=\frac{16-x}{5}\left(2\right)\)
MC = 30.
pt (2) <=>
\(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> 5(7x-1) + 60x = 6(16-x)
<=> 35x-5 + 60x = 96-6x
<=> 101x = 101
<=> x = 1
Vậy...
d. \(\frac{3x+2}{2}-\frac{3x+1}{6}=5\) (3)
MC = 12.
pt (3)<=>
\(\frac{6\left(3x+2\right)}{12}-\frac{2\left(3x+1\right)}{12}=\frac{60}{12}\)
=> 6(3x+2) - 2(3x+1) = 60
<=> 18x+12 - 6x-2 = 60
<=> 12x = 50
<=> x = \(\frac{25}{6}\)
Vậy...
e. \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) (4)
MC = 30.
pt (4) <=>
\(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> 6(x+4) - 30x + 120 = 10x - 15(x-2)
<=> 6x+24 - 30x + 120 = 10x - 15x+30
<=> -19x = -114
<=> x = \(\frac{114}{19}=6\)
Vậy...
d) \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x-2}{5}-5\)
\(\Leftrightarrow\frac{5\left(5x+2\right)}{30}-\frac{10\left(8x-1\right)}{30}=\frac{6\left(4x-2\right)}{30}-\frac{150}{30}\)
\(\Leftrightarrow25x+10-80x+10=24x-12-150\)
\(\Leftrightarrow25x-80x-24x=-12-150-10-10\)
\(\Leftrightarrow-79x=-182\)
\(\Leftrightarrow x=\frac{182}{79}\).
Vậy tập nghiệm phương trình \(s=\left\{\frac{182}{79}\right\}\)
a)\(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\)
\(\Leftrightarrow\frac{3\left(3x+2\right)}{6}-\frac{3x+1}{6}=\frac{10}{6}+\frac{12x}{6}\)
\(\Leftrightarrow9x+6-3x+1=10+12x\)
\(\Leftrightarrow9x-3x-12x=10-6-1\)
\(\Leftrightarrow-6x=3\)
\(\Leftrightarrow x=\frac{-1}{2}\).
Vậy tập nghiệm phương trình \(S=\left\{\frac{-1}{2}\right\}\)