1. (2 mũ x + 1)mũ 2 = 25
2. (x + 6) . (5 mũ x - 1) = 0
3. 2 . 3 mũ x + 3 mũ 2 + x = 891
4. (x - 3) mũ 2023 = x - 3
cứu em với ạ =(
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) 02002 < 02023
b) 20220 = 20230
c) 549 < 5510
d) ( 4 + 5 )3 > 42 + 52
đ) 92 - 32 > ( 9 - 3 )2
Bài 2:
a) 32 x 43 - 32 + 333
= 9 x 64 - 9 + 333
= 576 - 9 + 333
= 567 + 333
= 900
b) 5 x 43 + 24 x 5 + 410
= 5 x 64 + 24 x 5 + 1
= 5 x ( 64 + 24 ) + 1
= 5 x 88 + 1
= 440 + 1
= 441
c) 23 x 42 + 32 x 5 - 40 x 12023
= 8 x 16 + 9 x 5 - 40 x 1
= 128 + 45 - 40
= 133
Bài 1 :
a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)
b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)
c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)
d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)
đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
Lời giải :
1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)
\(=\frac{a^3}{4}+3ab^2\)
Lời giải :
2. \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy...
`(2^x+1)^2 =25`
`=> (2^x+1)^2 = (+-5)^2`
\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)
\(\left(x+6\right)\left(5^x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
\(\left(x-3\right)^{2023}=x-3\)
\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)