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a) \(2x\left(x-3\right)+6\left(3x-3\right)=0\)
\(\Leftrightarrow2x^2-6x+18x-18=0\)
\(\Leftrightarrow2x^2+12x-18=0\)
Mà \(2x^2\ge0\)
\(\Rightarrow x\in\varnothing\)
a)=>2x^2-6x+18x-18=0 b)=>6x^2-15x-75-30x =????
=>2x^2+12x=0+18
=>2x^2+12x=18
=>x.(2x+12)=18 (tự làm phần còn lai)
a. \(\left(\frac{-1}{3}\right)^3.x=\frac{1}{81}\)
\(\Leftrightarrow\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=\frac{-1}{3}\)
b. x8 = 16 . x6
<=> x8 : x6 = 16
<=> x2 = 42
<=> x = 4
c. (2x - 1)6 = (2x - 1)8
<=> x = \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy x = 1 hoặc 0
\(\frac{81}{3^{2x}+1}=3\)
=> 3.(32x + 1) = 81
=> 32x+1 = 27
=> 32x+1 = 33
=> 2x + 1 = 3
=> 2x = 2
=> x = 1
Đúng 100%
\(\left|x-3\right|=\left|2x+15\right|\)
\(\Rightarrow\orbr{\begin{cases}2x+15=x-3\\2x+15=3-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-18\\x=-4\end{cases}}}\)
KL:..................................................................
\(\left|2x-4\right|+\left|2x-6\right|=2\)
\(\Rightarrow\left|2x-4\right|+\left|6-2x\right|=2\)
Ta có: \(\hept{\begin{cases}\left|2x-4\right|\ge2x-4\forall x\\\left|6-2x\right|\ge6-2x\forall x\end{cases}\Rightarrow\left|2x-4\right|+\left|6-2x\right|\ge2x-4+6-2x=2\forall x}\)
Mà \(\left|2x-4\right|+\left|2x-6\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|2x-4\right|=2x-4\\\left|6-2x\right|=6-2x\end{cases}\Leftrightarrow\hept{\begin{cases}2x-4\ge0\\6-2x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge2\\x\le3\end{cases}\Rightarrow}2\le x\le3}\)
KL:.............................................................
Câu c tương tự câu b
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
\(a)x+\left(-5\right)=-14\)
\(\Leftrightarrow x=-14-\left(-5\right)\)
\(\Leftrightarrow x=-14+5\)
\(\Leftrightarrow x=-9\)
\(b)-x+7=-23\)
\(\Leftrightarrow-x=-23+ \left(-7\right)\)
\(\Leftrightarrow-x=-30\)
\(\Leftrightarrow x=30\)
\(c)112-x=\left(-3\right).\left(-15\right)\)
\(\Leftrightarrow112-x=45\)
\(\Leftrightarrow x=112-45\)
\(\Leftrightarrow x=67\)
\(d)\left(x-15\right)-27=5^5:5^3\)
\(\Leftrightarrow\left(x-15\right)-27=5^2\)
\(\Leftrightarrow\left(x-15\right)-27=25\)
\(\Leftrightarrow x-15=52\)
\(\Leftrightarrow x=67\)
\(e)\left(2x+1\right)^2=81\)
\(\Leftrightarrow\left(2x+1\right)^2=9^2\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
\(f)(x-5^3)=-27\)
\(f)(x-5^3)=-9^3\)
\(\Leftrightarrow x-5=-9\)
\(\Leftrightarrow x=-4\)
P/s: Bạn tự kết luận.
3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn
\(3x+\frac{3}{6}=7x+\frac{3}{15}\)
\(\Rightarrow4x=\frac{3}{10}\Rightarrow x=\frac{3}{40}\)
Ta co:
2x+3/6 = 7x + 3/15
=> 3/6 = 5x + 3/15
=> x = (3/6 - 3/15):5 = 3/50
\(3^{2x}\cdot3^{x+6}=81^{15}\)
\(3^{2x+x+6}=\left(3^4\right)^{15}\)
\(3^{3x+6}=3^{60}\)
\(3x+6=60\)
\(3x=60-6=54\)
\(x=54:3\)
\(x=18\)
32x . 3x + 6 = (34)15
=>32x+1 = 360
=>2x + 1 = 60
=> x = 59/2
KL