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a) (3x - 1)6 = (3x - 1)4
=> (3x - 1)6 - (3x - 1)4 = 0
=> (3x - 1)4 [(3x - 1)2 - 1] = 0
=> (3x - 1)4 = 0 hoặc (3x - 1)2 - 1 = 0
=> 3x - 1 = 0 hoặc (3x - 1)2 = 1
+) 3x - 1 = 0 => x = 1/3
+) 3x - 1 = 1 hoặc 3x - 1 = -1
=> 3x = 2 hoặc 3x = 0
=> x = 2/3 hoặc x = 0
Vậy x = 1/3,x = 2/3,x = 0
b) 5 - |3x - 1| = 3
=> |3x - 1| = 2
=> 3x - 1 = 2 hoặc 3x - 1 = -2
=> x = 1 hoặc x = -1/3
c) 9(x - 1)2 - 4/9 : 2/9 = 1/4
=> 9(x - 1)2 - 4/9 . 9/2 = 1/4
=> 9(x - 1)2 - 2 = 1/4
=> 9(x - 1)2 = 9/4
=> (x - 1)2 = 1/4
=> x - 1 = 1/2 hoặc x - 1 = -1/2
=> x = 3/2 hoặc x = 1/2
a) \(\left(3x-1\right)^6=\left(3x-1\right)^4\)
=>\(\hept{\begin{cases}3x-1=0\\3x-1=1\\3x-1=-1\end{cases}}=>\hept{\begin{cases}3x=1\\3x=2\\3x=0\end{cases}=>\hept{\begin{cases}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{cases}}}\)
Vậy x = \(\frac{1}{3}\);x=\(\frac{2}{3}\);x=0
\(4.3^x+3^{x+1}=63\)
\(\Rightarrow4.3^x+3.3^x=63\)
\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)
\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)
mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)
\(\Rightarrow x\in\varnothing\)
a) \(\left(x-\dfrac{4}{9}\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-\dfrac{4}{9}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{4}{9}=\dfrac{1}{2}\\x-\dfrac{4}{9}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{9}+\dfrac{1}{2}\\x=-\dfrac{1}{2}+\dfrac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{18}\\x=-\dfrac{1}{18}\end{matrix}\right.\)
b) Mình chưa rõ đề bài cho lắm
a/ 3x(x+1) - 2x(x+2) = -1 - x
<=> 3x2 + 3x -2x2 - 4x + x + 1 = 0
<=> x2 + 1 = 0 <=> x2 = -1 (vô lí)
vậy k có x nào t/m đề
b/ 4(x+2) - 7(2x-1) + 9(3x-4) = 30
<=> 4x + 8 - 14x + 7 + 27x - 36 = 30
<=> 17x = 30 + 36 - 7 - 8 = 51
<=> x = 51/17 = 3
Vậy x = 3
a/ 3x(x+1) - 2x(x+2) = -1 - x
<=> 3x2 + 3x -2x2 - 4x + x + 1 = 0
<=> x2 + 1 = 0
<=> x2 = -1 (vô lí)
vậy ko có x nào t/m đề ra
b/ 4(x+2) - 7(2x-1) + 9(3x-4) = 30
<=> 4x + 8 - 14x + 7 + 27x - 36 = 30
<=> 17x = 30 + 36 - 7 - 8 = 51
<=> x = \(\dfrac{51}{13}\) = 3
Vậy x = 3
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a) \(\left|x+9\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
e) \(\left|2x+4\right|=-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)
h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)
a) |9+x|=2x
TH1: 9+x=2x
<=> 9=2x-x
<=> x=9
TH2: -9-x=2x
<=> -9=3x
<=> x=-3
b) |5x|-3x=2
TH1: 5x-3x=2
<=> 2x=2
<=> x=1
TH2: -5x-3x=2
<=> -8x=2
<=>x=-4
c) |x+6|-9=2x
TH1: x+6-9=2x
<=> -3=x
TH2: -x-6-9=2x
<=> -15=3x
<=>x=-5
d) |2x-3|+x=21
TH1: 2x-3+x=21
<=> 3x=24
<=> x=8
TH2: -2x+3+x=21
<=> -x=18
<=> x=-18
e,i,g,h tương tự
\(2\left(x+4\right)-\dfrac{2}{3}x=9\)
\(\Leftrightarrow x\cdot\dfrac{4}{3}=1\)
hay \(x=\dfrac{3}{4}\)