K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 8 2017
\(x^2-36=0\Rightarrow x^2=36\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(3x^2-75=0\)
\(\Rightarrow3\left(x^2-25\right)=0\)
\(\Rightarrow x^2-25=0\Rightarrow x^2=25\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(4x^2-4x+1=0\)
\(\Rightarrow\left(2x-1\right)^2=0\)
\(\Rightarrow2x-1=0\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
\(\left(x+3\right)^2-4=0\)
\(\Rightarrow\left(x+3\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
19 tháng 8 2017
a) \(x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{36}\\x=-\sqrt{36}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
vậy \(x=6;x=-6\)
b) \(3x^2-75=0\Leftrightarrow3\left(x^2-25\right)=0\Leftrightarrow x^2-25=0\Leftrightarrow x^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{25}\\x=-\sqrt{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\) vậy \(x=5;x=-5\)
c) \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
d) \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x+3=\sqrt{4}\\x+3=-\sqrt{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\) vậy \(x=-1;x=-5\)
HH
1 tháng 8 2018
a/ \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)
b/ \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)
<=> \(\left(x-3\right)\left(x+19\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)
d/ \(4x^2-12x+9=0\)
<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)
<=> \(\left(2x-3\right)^2=0\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\)
13 tháng 10 2018
4x(x-2005)-(x+2005)=0
4x(x-2005)+(x-2005)=0
(x-2005)(4x+1)=0
<=>x-2005=>x=2005
4x+1=0=>x=-1/4
b, (x+1)2-x-1=0
(x+1)2-(x+1)=0
(x+1)(x+1-1)=0
(x+1)x=0
<=>x+1=0=>x=-1
x =0
21 tháng 7 2017
a) \(\left(x+17\right).\left(25-x\right)=0\)
\(\Leftrightarrow x+17=0\)hoặc \(25-x=0\)
Từ \(x+17=0\Rightarrow x=0-17=-17\)
Từ \(25-x=0\Rightarrow x=25-0=25\)
Vậy \(x=-17\) hoặc \(25\)
TX
1
26 tháng 7 2017
a)\(x^2-2x-24=0\Leftrightarrow x^2-2x+1-25=0\)
\(\Leftrightarrow\left(x-1\right)^2-5^2=0\Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
b)\(x^2+8x+12=0\Leftrightarrow x^2+8x+16-4=0\)
\(\Leftrightarrow\left(x+4\right)^2-2^2=0\Leftrightarrow\left(x+4-2\right)\left(x+4+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\Leftrightarrow\hept{\begin{cases}x=-2\\x=-6\end{cases}}\)
c)\(4x^2+4x-63=0\Leftrightarrow4x^2+4x+1-64=0\)
\(\Leftrightarrow\left(2x+1\right)^2-8^2=0\Leftrightarrow\left(2x+1-8\right)\left(2x+1+8=0\right)\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+9\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
1 tháng 9 2020
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
29 tháng 10 2017
a) ( 4x - 1 ) ( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{4};2\right\}\)
b) 4x2 - 12x = 0
<=> 4x ( x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x\in\left\{0;3\right\}\)
c) ( x - 5 )4 + 25 - x2 = 0
( x - 5 ) 4 + ( 5 - x ) ( 5 + x ) = 0
( x - 5 ) ( 4 + 5 + x ) = 0
( x - 5 ) ( 9 + x ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\9+x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-9\end{cases}}\)
Vậy \(x\in\left\{-9;5\right\}\)
\(4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}=3\\x=-\dfrac{6}{2}=-3\end{matrix}\right.\)
\(4x^2-36=0\)
\(4x^2\) \(=0+36\)
\(4x^2\) \(=36\)
\(x^2\) \(=36:4\)
\(x^2\) \(=9\)
\(x^2\) \(=3^2\)
\(=>x=3\)
Vậy...