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mk chỉ bn bn chỉ cần tính trong ngoặc ra bao nhiêu thì viết mũ rồi đổi tính bình thường nha
2x:2=32
==> 2x—1=25
==> x—1=5
x=5+1
x=6
5x—1:5=53
==> 5x—2=53
==> x—2=3
x—2=3
x=3+2
x=5
(2x—1)3=125
(2x—1)3=53
==> 2x—1=5
2x=5+1
2x=6
x=6:2
x=3
x17=x3
==>x=0 hoặc x=1
Mình quên cách lập luận bài này rồi bạn lên mạng tham khảo thêm nha
a) 2^x;2=32
Suy ra:2^x=32:2
Suy ra :2^x=16
Mà 16=2^4
Suy ra :x=4
Vậy x=4
Lát nữa mình giải nốt,bây giờ mình có việc.k cho mình nhé
a, \(\left(x-1\right)^3=125\)
\(\Rightarrow x-1=5\Rightarrow x=6\)
b, \(\left(2x+1\right)^3=343\)
\(\Rightarrow2x+1=7\Rightarrow x=3\)
c, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow7x-11=10\Rightarrow x=3\)
Chúc bạn học tốt!!!
a) \(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Rightarrow x-1=5\)
\(\Leftrightarrow x=6\) vậy \(x=6\)
b) \(\left(2x+1\right)^3=343\Leftrightarrow\left(2x+1\right)^3=7^3\Rightarrow2x+1=7\)
\(\Leftrightarrow2x=6\Leftrightarrow x=\dfrac{6}{2}=3\) vậy \(x=3\)
c) \(\left(7x-11\right)^3=2^5.5^2+200\Leftrightarrow\left(7x-11\right)^3=32.25+200\)
\(\Leftrightarrow\left(7x-11\right)^3=800+200=1000=10^3\Rightarrow7x-11=10\)
\(\Leftrightarrow7x=21\Leftrightarrow x=\dfrac{21}{7}=3\) vậy \(x=3\)
a) 3*2x =48 \(\Leftrightarrow\)2x=16 \(\Leftrightarrow\)x=4 b) (2x +1 )3 =53 \(\Leftrightarrow\)2x +1 =5 \(\Leftrightarrow\)2x =4 \(\Leftrightarrow\)x=2 c) 36+x=36 \(\Leftrightarrow\)x=0 d) 1+x-15=27-1 \(\Leftrightarrow\)x=40
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\) \(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=2+1=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
Giờ mới đúng thật nè
\(2\dfrac{1}{3}-125\%.x=\dfrac{1}{4}\)
⇒ \(\dfrac{7}{3}-\dfrac{125}{100}.x=\dfrac{1}{4}\)
⇒ \(\dfrac{5}{4}x=\dfrac{28}{12}-\dfrac{3}{12}\)
⇒ \(\dfrac{5}{4}x=\dfrac{25}{12}\)
⇒ \(x=\dfrac{25}{12}.\dfrac{4}{5}\)
⇒ \(x=\dfrac{100}{60}\) \(=\dfrac{5}{3}\)
Vậy \(x=\dfrac{5}{3}\).
\(2\dfrac{1}{3}\) - 125% . x = \(\dfrac{1}{4}\)
=> \(\dfrac{7}{3}\) - \(\dfrac{125}{100}\) . x = \(\dfrac{1}{4}\)
=> \(\dfrac{5}{4}x\) = \(\dfrac{28}{12}\) - \(\dfrac{3}{12}\)
=> \(\dfrac{5}{4}x\) = \(\dfrac{25}{12}\)
=> \(\dfrac{25}{12}\). \(\dfrac{4}{5}\)
=> \(\dfrac{100}{60}\)\(=\dfrac{5}{3}\)
x = \(\dfrac{5}{3}\)
a) (3x - 1)2 = 100
(3x - 1)2 = 102
=>3x - 1 = 10
=> 3x = 10 + 1
3x = 11
x = 11/3
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(1999^{2x-6}=1\)
\(\Rightarrow1999^{2x-1}=1999^0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
c) \(x^{2002}=x\)
\(\Rightarrow x^{2002}-x=0\)
\(\Rightarrow x.\left(x^{2001}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)
+) \(x=0\)
+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d) \(\left(x-1\right)^2=9\)
\(\Rightarrow x-1=\pm3\)
+) \(x-1=3\Rightarrow x=4\)
+) \(x-1=-3\Rightarrow x=-2\)
Vậy \(x\in\left\{4;-2\right\}\)
e) \(\left(2x-3\right)^2=81\)
\(\Rightarrow2x-3=\pm9\)
+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)
+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)
Vậy \(x\in\left\{6;-3\right\}\)
Các phần khác làm tương tự
\(\left(3x-1\right)^3=125\)
=>\(\left(3x-1\right)^3=5^3\)
=>3x-1=5
=>3x=6
=>\(x=\dfrac{6}{3}=2\)
\(\left(3x-1\right)^3=125\\ \Rightarrow\left(3x-1\right)^3=5^3\\ \Rightarrow3x-1=5\\ \Rightarrow3x=5+1\\ \Rightarrow3x=6\\ \Rightarrow x=6:3\\ \Rightarrow x=2\)