giúp tớ vs tớ cần gấp ạ :33
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A) Do 34 < 36 nên 34³⁴ < 34³⁶
B) 1²⁰²³ = 1
2023⁰ = 1
Vậy 1²⁰²³ = 2023⁰
C) Do 45 < 47 nên 45²⁰²³ < 47²⁰²³
Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) + 2 + \(\dfrac{5}{7}\)
= 2
2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8
= (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8
= \(\dfrac{2}{7}\) x 28 - 8
= 8 - 8
= 0
Đề bài ko chính xác, nếu x bất kì thì tồn tại vô số x để P nguyên
Nếu \(x\) nguyên thì mới có hữu hạn giá trị x
\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)
\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)
\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)
\(A=1-\dfrac{1}{23}\)
\(A=\dfrac{22}{23}\)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{10}\)
\(B=\dfrac{5}{10}-\dfrac{1}{10}\)
\(B=\dfrac{4}{10}\)
\(B=\dfrac{2}{5}\)
a, 27.\(3^x\) = 243
\(3^x\) = 243 : 27
\(3^x\) = 9
\(3^x\) = 32
\(x\) = 2
b, 49.7\(^x\) = 2041
7\(^x\) = 2041: 49
7\(^x\) = 41,65
\(x\) = log741,56
\(x\) ≈ 1,916
c, 64.4\(^x\) = 4\(^5\)
4\(^x\) = 45 : 64
4\(^x\) = 4
\(x\) = 1
d, 3\(^x\) = 243
3\(^x\) = 35
\(x\) = 5
e, 2\(^{x+5}\) = 128
\(2^{x+5}\) = 27
\(x+5\) = 7
\(x\) = 7 - 5
\(x\) = 2
f, 3.3\(^x\) = 81
3\(^x\) = 81 : 3
3\(^x\) = 27
3\(^x\) = 33
\(x\) = 3
g, 25 + 5\(^x\).5\(^x\) = 650
5\(^{2x}\) = 650 - 25
5\(^{2x}\) = 625
5\(^{2x}\) = 54
2\(x\) = 4
\(x\) = 2
a: Kẻ Ox//AB
Ox//AB
=>góc xOA=góc OAB(hai góc so le trong)
=>góc xOA=41 độ
góc xOA+góc xOB=góc AOB
=>góc xOB=71-41=30 độ=góc OCD
=>Ox//CD
=>AB//CD
=>Ax//Cy
b: BD//AO
=>góc B+góc OAB=180 độ(trong cùng phía)
=>góc B=180-41=139 độ
AB//CD
=>góc B+góc D=180 độ(hai góc trong cùng phía)
=>góc D=180-139=41 độ
Tính được các góc \(\widehat{BAC}=80^{\circ};\widehat{MNP}=40^{\circ};\widehat{QSR}=80^{\circ};\widehat{IHK}=40^\circ\)
- Vì \(\widehat{B}=\widehat{P};\widehat{C}=\widehat{N};BC=NP\) nên \(\Delta ABC=\Delta MPN\quad\left(g.c.g\right)\)
Tương tự, \(\Delta ABC=\Delta FED\quad\left(g.c.g\right);\Delta MPN=\Delta FED\quad\left(g.c.g\right)\)
- Vì \(\widehat{Q}=\widehat{K};\widehat{S}=\widehat{I};QS=IK\) nên \(\Delta QSR=\Delta KHI\left(g.c.g\right)\)