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CN
22 tháng 12 2017
a) \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b) (2x - 1)2 - (x + 3)2 = 0
⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0
⇔ (x - 4)(3x +2) = 0
⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)
c) 2x2 - x - 6 = 0
⇔ 2x2 - 4x + 3x - 6 = 0
⇔ 2x(x - 2) + 3(x - 2) = 0
⇔ (x - 2) (2x + 3) = 0
⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)
NT
22 tháng 12 2017
2)a.
\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)
b)
\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
13 tháng 9 2018
a) \(x^2-16=0\Rightarrow x^2=16\Rightarrow x^2=\pm4\)
b) \(4x^2-9=0\Rightarrow\left(2x-3\right)\left(2x+3\right)=0\Rightarrow x=\pm1,5\)
c) \(25x^2-1=0\Rightarrow\left(5x-1\right)\left(5x+1\right)=0\Rightarrow x=\pm0,2\)
d) \(4\left(x-1\right)^2-9=0\Rightarrow\left(2x-2-3\right)\left(2x-2+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-5=0\Rightarrow x=2,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(25x^2-\left(5x+1\right)^2=0\Rightarrow\left(5x+5x+1\right)\left(5x-5x-1\right)=0\Rightarrow10x+1=0\Rightarrow x=-0,1\)
f) \(\dfrac{1}{4}-9\left(x-1\right)^2=0\Rightarrow\left(\dfrac{1}{2}+3x-3\right)\left(\dfrac{1}{2}-3x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{7}{6}\end{matrix}\right.\)
g) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\Rightarrow\left(\dfrac{1}{4}+2x+\dfrac{3}{4}\right)\left(\dfrac{1}{4}-2x-\dfrac{3}{4}\right)=0\Rightarrow\left[{}\begin{matrix}x=-0,5\\x=-0,25\end{matrix}\right.\)
h) \(\dfrac{1}{9}x^2-\dfrac{2}{3}x+1=0\Rightarrow\left(\dfrac{1}{3}x-1\right)^2=0\Rightarrow\dfrac{1}{3}x=1\Rightarrow x=3\)
k) \(4\left(x-3\right)^2-\left(2-3x\right)^2=0\Rightarrow\left(2x-6+2-3x\right)\left(2x-6-2+3x\right)=0\Rightarrow\left[{}\begin{matrix}-x-4=0\Rightarrow x=-4\\5x-8=0\Rightarrow x=1,6\end{matrix}\right.\)
l) \(x^2-x-12=0\Rightarrow x^2-4x+3x-12=0\Rightarrow x\left(x-4\right)+3\left(x-4\right)=0\Rightarrow\left(x+3\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
23 tháng 9 2016
a) \(x+5x^2=0\)
\(=>x\left(1+5x\right)=0\)
\(=>\hept{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
b) \(x^3+x=0\)
\(=>x\left(x^2+1\right)=0\)
\(=>\hept{\begin{cases}x=0\\x^2+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x\in\phi\end{cases}}\)
c) \(5x\left(x-1\right)=x-1\)
\(=>5x\left(x-1\right)-x+1=0\)
\(=>5x\left(x-1\right)-\left(x-1\right)=0\)
\(=>\left(x-1\right)\left(5x-1\right)=0\)
\(=>\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-10x=-25\)
\(=>x^2-10x+25=0\)
\(=>\left(x-5\right)^2=0\)
\(=>x-5=0\)
\(=>x=5\)
23 tháng 9 2016
\(a,x+5x^2=0\)
\(x.\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
31 tháng 7 2022
b: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
c: \(\Leftrightarrow x\left(x^2-4x+5\right)=0\)
=>x=0
d: \(\Leftrightarrow2\cdot2^x-10\cdot2^x=-16\)
\(\Leftrightarrow-8\cdot2^x=-16\)
\(\Leftrightarrow2^x=2\)
hay x=1
DN
5 tháng 9 2018
\(a,x^2-2x=0\)
\(\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy ...
\(b,\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2=16\)
\(\Rightarrow\left(5-2x\right)^2=4^2\)
\(\Rightarrow5-2x=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{9}\end{matrix}\right.\)
Vậy ...
\(c,x\left(x+3\right)-x^2-11=0\)
\(\Rightarrow x^2+3x-x^2-11=0\)
\(\Rightarrow3x-11=0\)
\(\Rightarrow3x=11\)
\(\Rightarrow x=\dfrac{11}{3}\)
Vậy ...
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
29 tháng 11 2022
a: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-16}{16\left(x^2+x+1\right)}\cdot\left(x+1\right)=-\dfrac{x+1}{x^2+x+1}\)
b: \(B=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x+2}{x^2+x+1}\)
\(P=A+B=\dfrac{-x-1+x+2}{x^2+x+1}=\dfrac{1}{x^2+x+1}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< =1:\dfrac{3}{4}=\dfrac{4}{3}\)
Dấu = xảy ra khi x=-1/2
a:Ta có: \(16-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)