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Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a: =>|5/4x-7/2|=|5/8x+3/5|
=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5
=>5/8x=41/10 hoặc 15/8x=29/10
=>x=164/25 hoặc x=116/75
b: =>3:|x/4-2/3|=6-21/5=9/5
=>|1/4x-2/3|=5/3
=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3
=>1/4x=7/3 hoặc 1/4x=-1
=>x=28/3 hoặc x=-4
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)
e: =>|2x-7|=2x-7
=>2x-7>=0
=>x>=7/2
b) \(\left|2x-5\right|\)= x+1
\(\Rightarrow\) 2x-5 = x+1
\(\Rightarrow\) 2x-x=1+5
\(\Rightarrow\) x = 6
c) \(\left|3x-2\right|\)-1= x
\(\Rightarrow\) 3x-2-1= x
\(\Rightarrow\)3x-x =2+1
\(\Rightarrow\)2x =3
\(\Rightarrow\) x =\(\dfrac{3}{2}\)=1,5
e) \(\left|7-2x\right|\)+7 = 2x
\(\Rightarrow\)7-2x+7 =2x
\(\Rightarrow\) -2x -2x = -7-7
\(\Rightarrow\) -4x = -14
\(\Rightarrow\) x=\(\dfrac{14}{4}\)=\(\dfrac{7}{2}\)
\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{7};5\right\}\)
\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)
\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
Giải tiếp tương tự
Sau đó giải tiếp câu còn lại
\(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Leftrightarrow\left(\frac{\left(5x-1\right)\left(2x-1\right)}{3}\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x=\frac{1}{5}\)hoặc\(x=\frac{1}{2}\)
a. ( 2x - 5) ( x -3 ) = \(2x^2\)
=> \(2x^2-6x-5x+15\) = \(^{ }2x^2\)
=> \(2x^2-2x^2-6x-5x=-15\)
=> -11x = -15
=> x = \(\dfrac{15}{11}\)
b. (-2x+1)(4x-1)=(7-x).8x
=> \(^{ }-8x^2+2x+4x-1=56x-8x^2\)
=> \(^{ }-8x^2+8x^2+2x+4x-56x=1\)
=> -50x = 1
=> x = \(\dfrac{-1}{50}\)
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;3\right\}\)
b: \(\Leftrightarrow\left|4x+1\right|=8x-x-2=7x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(7x-2-4x-1\right)\left(7x-2+4x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(3x-3\right)\left(11x-1\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
c: |17x-5|=|17x+5|
=>17x-5=17x+5 hoặc 17x+5=5-17x
=>x=0