Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
\(b,x^6=x^2\)
\(x^6-x^2=0\)
\(x^2\cdot\left(x^4-1\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(c\text{}\text{}\text{}\text{},\left(x-2\right)\cdot\left(x-5\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
\(d,x^{10}-x^5=0\)
\(x^5\cdot\left(x^5-1\right)=0\)
\(\orbr{\begin{cases}x^5=0\\x^5=1\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
\(e,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\left(x-5\right)^4\cdot\left[1-\left(x-5\right)^2\right]=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm1+5\end{cases}}}\)
\(\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
\(\left(2x+1\right)^3=125\Rightarrow\left(2x+1\right)^3==5^3\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\Rightarrow x=4:2=2\)
\(x^6=x^2\Rightarrow x^2.x^4=x^2\)Vì vậy nên \(x=\pm1\)
\(\left(x-2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\Rightarrow x=0+2=5\\x-5=0\Rightarrow X=0+5=5\end{cases}}\)
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
\(c,-5\left(2-x\right)+4\left(x-3\right)=10x-15\)
\(-10+5x+4x-12=10x-15\)
\(5x+4x-10x-10-12=-15\)
\(-x-10=-3\)
\(-x=7\)
\(x=-7\)
Vậy \(x=-7.\)
Những câu khác bạn nên tự làm...
a) −7. (5 - x) − 2. (x − 10) = 15
-35 + 7x - 2x + 20 = 15
7x - 2x = 35 - 20 + 15
5x = 30
x = 30 : 5
x = 6
Vậy x = 6
b) 4. (2 − x) + 3(x − 5) = 14
8 - 4x + 3x - 15 = 14
8 - 15 - 14 = 4x - 3x
-21 = x
Vậy x = -21
c) −5. (2 − x) + 4. (x − 3) = 10x −15
-10 + 5x + 4x - 12 = 10x - 15
-10 - 12 + 15 = 10x - 5x - 4x
-7 = x
Vậy x = -7
d) −7. (3x − 5) + 2. (7x − 14) = 28
-21x + 35 + 14x - 28 = 28
35 - 28 - 28 = 21x - 14x
-21 = 7x
x = (-21) : 7
x = -3
Vậy x = -3
e) 5. (4 − x) − 7(−x + 2) = 4 − 9 + 3
20 - 5x + 7x - 14 = -2
-5x + 7x = -2 - 20 + 14
2x = -8
x = (-8) : 2
x = -4
Vậy x = -4
f) 5. (x − 7) + 10. (3 − x) = 20
5x - 35 + 30 - 10x = 20
-35 + 30 - 20 = -5x + 10x
-25 = 5x
x = (-25) : 5
x = -5
Vậy x = -5
g) −4. (x + 1) + 8. (x − 3) = 24
-4x - 4 + 8x - 24 = 24
-4x + 8x = 24 + 4 + 24
4x = 52
x = 52 : 4
x = 13
Vậy x = 13
h) 4. (x − 1) − 3. (x − 2) = −|−5|
4x - 4 - 3x + 6 = -5
4x - 3x = -5 + 4 - 6
x = -7
Vậy x = -7
\(3.\left(x-2\right)+1=25\Leftrightarrow\left(x-2\right)=8\Leftrightarrow x=10\)