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\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
\(b,x^6=x^2\)
\(x^6-x^2=0\)
\(x^2\cdot\left(x^4-1\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(c\text{}\text{}\text{}\text{},\left(x-2\right)\cdot\left(x-5\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
\(d,x^{10}-x^5=0\)
\(x^5\cdot\left(x^5-1\right)=0\)
\(\orbr{\begin{cases}x^5=0\\x^5=1\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
\(e,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\left(x-5\right)^4\cdot\left[1-\left(x-5\right)^2\right]=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm1+5\end{cases}}}\)
\(\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
\(\left(2x+1\right)^3=125\Rightarrow\left(2x+1\right)^3==5^3\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\Rightarrow x=4:2=2\)
\(x^6=x^2\Rightarrow x^2.x^4=x^2\)Vì vậy nên \(x=\pm1\)
\(\left(x-2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\Rightarrow x=0+2=5\\x-5=0\Rightarrow X=0+5=5\end{cases}}\)
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,5^x=125
=>5^x=5^3
=>x=3
b,3^2x=81
=>3^2x=3^4
=>2x=4
=>x=4:2=2
c,5^2x-3-2*5^2=5^2+3
5^2x-3-50=75
5^2x-3=75+50=125
5^2x-3=5^3
=>2x-3=3
=>2x=3+3=6
=>6:2=3
k cho mk nhé
\(a,125=5\cdot5\cdot5=5^3\Leftrightarrow x=3\)
\(b,81=3\cdot3\cdot3\cdot3=3^4\Leftrightarrow2x=4\Leftrightarrow x=4:2\Leftrightarrow x=2\)
\(c,5^{2x-3}-2\cdot5^2=5^2\cdot3\)
\(\Leftrightarrow5^{2x-3}=2\cdot5^2+5^2\cdot3\)
\(\Leftrightarrow5^{2x-3}=5^2\cdot\left(2+3\right)\)
\(\Leftrightarrow5^{2x-3}=5^2\cdot5\Leftrightarrow5^{2x-3}=5^3\)
\(\Leftrightarrow2x-3=3\Leftrightarrow2x=3+3\Leftrightarrow2x=6\Leftrightarrow x=6:2\Leftrightarrow x=3\)
\(\left(5^{2x}\cdot5^{x+2}\right):25=125^2\)
\(5^{2x+x+2}=125^2\cdot25\)
\(5^{3x+2}=\left(5^3\right)^2\cdot5^2\)
\(5^{3x+2}=5^6\cdot5^2\)
\(5^{3x+2}=5^8\)
\(\Rightarrow3x+2=8\)
\(3x=8-2\)
\(3x=6\)
\(x=6:3\)
\(x=2\)
a x+35=515/5=103
x=103-35=68
b 3(x+1)=96-42=54
x+1=54/3=18
x=18-1=7
a) \(5\left(x+35\right)=515\)
\(\Rightarrow x+35=103\)
\(\Rightarrow x=68\)
b) \(96-3\left(x+1\right)=42\)
\(\Rightarrow3\left(x+1\right)=54\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
c) \(5^x.5=5^4\Rightarrow5^x=5^3\Rightarrow x=3\)
d) \(\left(x-1\right)^2=125\)
Mà \(\orbr{\begin{cases}\left(5\sqrt{5}\right)^2=125\\\left(-5\sqrt{5}\right)^2=125\end{cases}\Rightarrow\orbr{\begin{cases}x-1=5\sqrt{5}\\x-1=-5\sqrt{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\sqrt{5}+1\\x=1-5\sqrt{5}\end{cases}}}\)
Mà lớp 6 chưa học căn
=> Kiểm tra lại đề
a.
2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
x = 5
b.
x15 = x
Vậy x = 0 hoặc x = 1 hoặc x = -1
c.
(2x + 1)3 = 125
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
d.
(x - 5)4 = (x - 5)6
TH1:
x - 5 = 0
x = 5
TH2:
x - 5 = -1
x = -1 + 5
x = 4
TH2:
x - 5 = 1
x = 1 + 5
x = 6
Vậy x = 5 hoặc x = 4 hoặc x = 6
Chúc bạn học tốt ^^
a) \(2^x.4=128\)
=> \(2^x=32\) => \(2^x=2^5\) => x = 5
b) \(x^{15}=x\) => x = 1 hoặc x = 0
c) \(\left(2x+1\right)^3=125\)
=> \(\left(2x+1\right)^3=5^3\) => 2x + 1 = 5 => x = 2
d) \(\left(x-5\right)^4=\left(x-5\right)^6\)
=> x - 5 = 0 hoặc x - 5 = 1
=> x = 5 hoặc x= 6
Chúc bạn làm bài tốt
\(2^x.4=128\)
\(2^x=128:4\)
\(2^x=32\)
\(\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
\(x^{15}=x\Leftrightarrow x\in\left\{-1;0;1\right\}\)
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
\(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Leftrightarrow\hept{\begin{cases}x-5=-1\\x-5=0\\x-5=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\x=5\\x=6\end{cases}}\)
\(\text{Vậy:}\)\(x\in\left\{4;5;6\right\}\)
\(2^x.4=128\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5.\)
\(x^{15}=x\Rightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)
\(\left(2x+1\right)^3=125\)
<=> \(\left(2x+1\right)^3=5^3\)
<=> \(2x+1=5\)
<=> \(x=2\)
\(\left(x-5\right)^6=\left(x-5\right)^4\)
<=> \(\left(x-5\right)^6-\left(x-5\right)^4=0\)
<=> \(\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
Giải ra được x = 5 ; x = 6 ; x = 4 .
\(\left(x^{54}\right)^2=x^{54.2}=x^{108}\\ Mà:\left(x^{54}\right)^2=5^{108}\Leftrightarrow x^{108}=5^{108}=\left(-5\right)^{108}\\ Vậy:x=5.hoặc.x=-5\\ ---\\ \left(5^x\right)^2=125^3:5^2\\ \Leftrightarrow5^{2x}=\left(5^3\right)^3:5^2=5^9:5^2=5^7\\ Vậy:2x=7\\ Vậy:x=\dfrac{7}{2}\)
\(\left(x^{54}\right)^2=5^{108}\\ \Rightarrow x^{108}=5^{108}\)
\(\Rightarrow x=5.\)
\(\left(5^x\right)^2=125^3:5^2\\ \Rightarrow\left(5^x\right)^2=\left(5^3\right)^3:5^2\\ \Rightarrow\left(5^x\right)^2=5^9:5^2\\ \Rightarrow\left(5^x\right)^2=5^7\\ \Rightarrow x.2=5\\ \Rightarrow x=2,5\)