(6x-48).(72-9x)=0
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a) \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
Vậy...
a) 4x + 3x = 217
x( 4 + 3 ) = 217
7x = 217
x = 217 : 7 = 31
Vậy x = 31
b) 9x - 3x = 216
( 9 -3)x = 216
6x = 216
x = 216:6 = 36
Vậy x = 36
c) 6x - 3x + 23 = 230
( 6 - 3 )x = 230 - 23
3x = 207
x = 207 : 3 = 69
Vậy x = 69
d) 5x + 3x + x = 72
5x + 3x + 1x = 72
( 5 + 3 + 1 )x = 72
9x = 72
x = 72 : 9 = 8
Vậy x = 8
Chúc bạn học tốt nhé
a) \(4x+3x=217\)
\(\Rightarrow x\cdot\left(3+4\right)=217\)
\(\Rightarrow7x=217\)
\(\Rightarrow x=\dfrac{217}{7}\)
\(\Rightarrow x=31\)
b) \(9x-3x=216\)
\(\Rightarrow x\cdot\left(9-3\right)=216\)
\(\Rightarrow6x=216\)
\(\Rightarrow x=\dfrac{216}{6}\)
\(\Rightarrow x=36\)
c) \(6x-3x+23=230\)
\(\Rightarrow x\cdot\left(6-3\right)=230-23\)
\(\Rightarrow3x=207\)
\(\Rightarrow x=\dfrac{207}{3}\)
\(\Rightarrow x=69\)
d) \(5x+3x+x=72\)
\(\Rightarrow x\cdot\left(5+3+1\right)=72\)
\(\Rightarrow9x=72\)
\(\Rightarrow x=\dfrac{72}{9}\)
\(\Rightarrow x=8\)
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
\(x^3-9x+7x^2-63=0\)
\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)
\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)
Vậy ...
x3−9x+7x2−63=0x3−9x+7x2−63=0
⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0
⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0
⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0
⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7
Vậy ...
\(3x^4-48\)
\(=\left(3x^4-6x^3\right)+\left(6x^3-12x^2\right)+\left(12x^2-24x\right)+\left(24x-48\right)\)
\(=3x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+24\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(3x^3+6x^2\right)+\left(12x+24\right)\right]\)
\(=\left(x-2\right)\left[3x^2\left(x+2\right)+12\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(3x^2+12\right)\)
\(x^4-8x\)
\(=x\left(x^3-8\right)\)
\(=x\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(4x-8\right)\right]\)
\(=x\left[x^2\left(x-2\right)+2x\left(x-2\right)+4\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
(6x - 48)(79 - 9x) = 0
6x - 48 = 0 hoặc 79 - 9x = 0
*) 6x - 48 = 0
6x = 48
x = 48 : 6
x = 8
*) 79 - 9x = 0
9x = 79
x = 79/9
Vậy x = 8; x = 79/9
Ta có :
\(9x^2+6x+2=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=-1\)
\(\Rightarrow\) Phương trình vô nghiệm .
(3-9x)(6x-4)(5-15x)=0
⇔ 3-9x=0 -> x=1/3
6x-4=0 -> x= 2/3
5-15x=0 -> x= 1/3
Vậy tập nghiệm S={ 2/3;1/3}
(6x - 48)(72 - 9x) = 0
6x - 48 = 0 hoặc 72 - 9x = 0
*) 6x - 48 = 0
6x = 48
x = 48 : 6
x = 8
*) 72 - 9x = 0
9x = 72
x = 72 : 9
x = 8
Vậy x = 8
x=8,x=8