Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có " (x - 5)7 = (x - 5)4
=> (x - 5)7 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)3 - 1] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
\(\left|2x\right|+2x=0\)
\(\Rightarrow\left|2x\right|=-2x\)
\(\Rightarrow2x\le0\)
\(\Rightarrow x\le0\)
Vậy \(x\le0\)
\(\left(x-1\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
\(\left|x-3\right|+x-3=0\)
\(\left|x-3\right|=-x+3\)
\(\left|x-3\right|=-\left(x-3\right)\)
\(\Rightarrow x-3\le0\)
\(\Rightarrow x\le3\)
Vậy \(x\le3\)
\(\left(x+1\right)^3=\left(x+1\right)^5\)
\(\left(x+1\right)^5-\left(x+1\right)^3=0\)
\(\left(x+1\right)^3.\left[\left(x+1\right)^2-1\right]=0\)
\(\orbr{\begin{cases}\left(x+1\right)^3=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)hoặc \(x=-2\)
Vậy \(x\in\left\{-1;0;-2\right\}\)
\(\left(x-2\right)^3=2^9\)
\(\left(x-2\right)^3=\left(2^3\right)^3\)
\(\Rightarrow x-2=2^3\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
Câu 6 tương tự câu 4
Tham khảo nhé~
P/S: nên chia nhỏ đăng thành nhiều bài khác nhau
\(2^x=2\Rightarrow x=1\)
\(2^{2x+2}=8^2\Rightarrow2^{2x+2}=2^6\Rightarrow2x+2=6\)\(2x=6-2=3\Rightarrow x=3:2=\frac{3}{2}\)
ne ban minh biet cau tra loi nhung lam the nao bam ngoac vuong
\(3^{2x-4}-x^0=8\)
\(3^{2x-4}=9\)
\(3^{2x-4}=3^3\)
\(\Rightarrow2x-4=3\)
\(2x=7\)
\(x=3,5\)
\(3^{2.x-4}-x^0=8\)
\(3^{2.x}:3^4-1=8\)
\(3^{2.x}:81=8+1\)
\(3^{2.x}:81=9\)
\(3^{2.x}=9.81\)
\(3^{2.x}=729\)
\(3^{2.x}=3^6\)
\(\Rightarrow2.x=6\)
\(\Rightarrow x=6:2=3\)
Vậy x = 3
a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)
\(=>142-12x-30=10^2\)
\(112-12x=100\)
\(=>12x=112-100=12=>x=1\)
Vậy số cần tìm là 1;
b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)
\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)
\(=>5x=5.3^4=>x=3^4=81\)
Vậy x=81;
CHÚC BẠN HỌC TỐT.......
Ta có: \(71.2-6(2x+5)=10^5:10^3\)
\(\Rightarrow142-12x-30=10^2\)
\(\Rightarrow142-30-10^2=12x\)
\(\Rightarrow142-30-100=12x\)
\(\Rightarrow12=12x\)
\(\Rightarrow x=\dfrac{12}{12}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
3^2x-4 -x^0 = 8
(3^2)^x : 3^4 -1 = 8
9^x : 81 =8 + 1
9^x : 81 =9
9^x = 9 . 81
9^x = 729
=>x=3
vậy x=3
\(3^{2x-4}-x^0=8\)
\(\Leftrightarrow3^{2x-4}-1=8\)
\(\Leftrightarrow3^{2x-4}=9\)
\(\Leftrightarrow3^{2x-4}=3^2\)
\(\Rightarrow2x-4=2\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)