(21-3x).(2x-10).(6-2x=0
2011(2x-6.(21-3x))=1
33x+2 + 5.32x. 23+234= 4203
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a/\(-\left(-432-3x\right)+\left(-432-4x+21\right)=-2345\)
\(\Leftrightarrow+432+3x-432-4x+21=-2345\)\(\Leftrightarrow\left[432+\left(-432\right)\right]+\left[\left(-4x\right)+3x\right]+21=-2345\)
\(\Leftrightarrow0+-1x+21=-2345\)\(\Leftrightarrow-1x=-2345-21=-2366\)
\(\Rightarrow x=\left(-2366\right):\left(-1\right)=2366\)
b/\(12x-\left(-3546-3x\right)=\left(2x-234\right)-\left(-678-2x\right)\)
\(\Leftrightarrow12x+3546+3x=2x-234+678+2x\)
\(\Leftrightarrow15x+3546=\left(2x+2x\right)+\left[\left(-234\right)+678\right]\)
\(\Leftrightarrow15x+3546=4x+444\)
\(\Leftrightarrow15x+3546-4x-444=0\)
\(\Leftrightarrow\left(15x-4x\right)+\left(3546-444\right)=0\)
\(\Leftrightarrow11x+3102=0\)
\(\Leftrightarrow11x=0-3102=-3102\)
\(\Rightarrow x=\left(-3102\right):11=-282\)
a) 1 + 3 + 5 + 7 + ... + (2x - 1) = 225
(2x - 1 + 1) . [(2x - 1 - 1) : 2 + 1] : 2 = 225
2x.[(2x - 2) : 2 + 1] : 2 = 225
x . (x - 1 + 1) = 225
x . x = 225
=> x = 15
a, \(\left|9+x\right|=2x\)
\(\Leftrightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}\Leftrightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)
b, \(\left|5x\right|-3x=2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3x=2\\5x-3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
\(a,\)( sửa lại xíu đề cho đúng nhé )
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=-\frac{2x}{x^2+x+1}\)
\(\Rightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1-3x^2=-2x^2+2x\)
\(\Rightarrow x=1\)
\(g,\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=-16\)
\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=-16\)
Đặt \(x^2+10x+16=a\)
\(\Rightarrow a\left(a+8\right)=-16\)
\(\Rightarrow a^2+8a+16=0\)
\(\Rightarrow\left(a+4\right)^2=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Rightarrow x^2+10x+25-25=0\)
\(\Rightarrow\left(x+5\right)^2-\left(\sqrt{5}\right)^2=0\)
\(\Rightarrow\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)