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BT
0
17 tháng 8 2017
* \(x^2-8x+12=0\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\) vậy \(x=2;x=6\)
* \(x^2+5x-14=0\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\) vậy \(x=-7;x=2\)
* \(16x^2-81=0\Leftrightarrow16\left(x^2-\dfrac{81}{16}\right)=0\Leftrightarrow x^2-\dfrac{81}{16}=0\)
\(\Leftrightarrow x^2=\dfrac{81}{16}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{81}{16}}\\x=-\sqrt{\dfrac{81}{16}}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\) vậy \(x=\dfrac{9}{4};x=\dfrac{-9}{4}\)
17 tháng 8 2017
+ \(x^2-8x+12=0\)
\(\Rightarrow\left(x^2-2.4x+16\right)-4=0\)
\(\Rightarrow\left(x-4\right)^2-4=0\)
\(\Rightarrow\left(x-4\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
+ \(16x^2-81=0\)
\(\Rightarrow16x^2-9^2=0\)
\(\Rightarrow16x^2=9^2\)
\(\Rightarrow x^2=\dfrac{81}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{81}{16}}\\x=-\sqrt{\dfrac{81}{16}}\end{matrix}\right.\)
TC
0
NT
1
N
5 tháng 11 2018
\(a,9x^2-49=0\)
\(9x^2=49\)
\(x^2=\frac{49}{9}=\frac{7^2}{3^2}=\frac{\left(-7\right)^2}{\left(-3\right)^2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
vậy ...
\(c,x^3-16x=0\)
\(x.\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=4,x=-4\end{cases}}\)
vậy ...
HT
1
28 tháng 3 2017
<=> (x2 +x +4)2 + 2 . 4x(x2+ x + 4) + (4x)2 = 0
<=> ( x2 + x+ 4 +4x )2 = 0
<=> [(x2 + x) + (4 +4x)] =0
<=> [x(x+1) + 4(1+x)] =0
<=> (x+1) + (x+4) =0
- x+1 = 0 <=> x= -1
- x+4 = 0 <=> x= -4
24 tháng 2 2020
\(4+2x\left(2x+4\right)=-x\)
\(4+2x.2x+8x=-x\)
\(4x+8x+x=-4\)
\(13x=-4\)
\(x=-\frac{4}{13}\)
Vậy pt có nghiệm là { -4/13 }
AH
Akai Haruma
Giáo viên
26 tháng 9 2018
Lời giải:
a)
\(x^2-2x=24\)
\(\Leftrightarrow x^2-6x+4x-24=0\)
\(\Leftrightarrow x(x-6)+4(x-6)=0\Leftrightarrow (x+4)(x-6)=0\)
\(\Rightarrow \left[\begin{matrix} x+4=0\\ x-6=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-4\\ x=6\end{matrix}\right.\)
b)
\(x^3-7x+6=0\Leftrightarrow (x^3-x)-(6x-6)=0\)
\(\Leftrightarrow x(x^2-1)-6(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-6(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-6)=0\)
\(\Leftrightarrow (x-1)(x^2-2x+3x-6)=0\)
\(\Leftrightarrow (x-1)[x(x-2)+3(x-2)]=0\)
\(\Leftrightarrow (x-1)(x-2)(x+3)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=2\\ x=-3\end{matrix}\right.\)
c) Xem lại đề.
d) Đặt \(x^2+x+4=a\) thì pt trở thành:
\(a^2+8ax+16x^2=0\)
\(\Leftrightarrow a^2+2.a.4x+(4x)^2=0\)
\(\Leftrightarrow (a+4x)^2=0\Rightarrow a+4x=0\)
\(\Rightarrow x^2+x+4+4x=0\)
\(\Rightarrow x(x+1)+4(x+1)=0\Leftrightarrow (x+1)(x+4)=0\)
\(\Rightarrow \left[\begin{matrix} x+4=0\rightarrow x=-4\\ x+1=0\rightarrow x=-1\end{matrix}\right.\)
22 tháng 1 2019
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
đầu bài yêu cầu gì thế >?>