d)(x-3).(x2+12)=0
e) |x-7|=6
i)120.(1-x).(8+x)=0
j)|x-5| +12=24
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\(a)\left(x-5\right).2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
\(b)x.3-13=47\)
\(\Rightarrow x.3=47+13\)
\(\Rightarrow x.3=60\)
\(\Rightarrow x=60:3\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
\(c)\left(x.7-7\right)\left(x.12+24\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x.7-7=0\Rightarrow x.7=7\Rightarrow x=1\\x.12+24=0\Rightarrow x.12=-24\Rightarrow x=-2\end{cases}}\)
Vậy\(x\in\left\{1;-2\right\}\)
\(e)140-10.x=120\)
\(\Rightarrow10.x=140-120\)
\(\Rightarrow10.x=20\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(g)x.5-127=273\)
\(\Rightarrow x.5=273+127\)
\(\Rightarrow x.5=400\)
\(\Rightarrow x=400:5\)
\(\Rightarrow x=80\)
Vậy \(x=80\)
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2+12=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x\in\varnothing\end{cases}}\)
e) \(\left|x-7\right|=6\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)
i) \(120\left(1-x\right)\left(8+x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(8+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}1-x=0\\8+x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-8\end{cases}}\)
j) \(\left|x-5\right|+12=24\)
\(\Leftrightarrow\left|x-5\right|=12\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=12\\x-5=-12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=17\\x=-7\end{cases}}\)
Bài giải
d, \(\left(x-3\right)\left(x^2+12\right)=0\)
Mà \(x^2+12\ne0\) nên \(x-3=0\)
\(\Rightarrow\text{ }x=3\)
e, \(\left|x-7\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-7=-6\\x-7=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=13\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-1\text{ ; }13\right\}\)
i, \(120\cdot\left(1-x\right)\cdot\left(8+x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-x=0\\8+x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-8\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{1\text{ ; }-8\right\}\)
j, \(\left|x-5\right|+12=24\)
\(\left|x-5\right|=24-12\)
\(\left|x-5\right|=12\)
\(\Rightarrow\orbr{\begin{cases}x-5=-12\\x-5=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-7\\x=17\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-7\text{ ; }17\right\}\)