Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
a hơi dài để làm phần b trước :
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot3^2-2^n\cdot2^2+3^n-2^n\)
\(=\left(3^n\cdot3^2+3^n\right)-\left(2^n\cdot2^2+2^n\right)\)
\(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^3.3\right)^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{18}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5.\left(2^6-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(A=\frac{2}{3.\left(64-1\right)}-\frac{5.\left(-6\right)}{9}\)
\(A=\frac{2}{3.63}+\frac{30}{9}\)
Tự lm tiếp Ball nhé~
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^6.\left(3+1\right)}\)
\(=\frac{2^{12}.3^4.2}{2^{12}.3^6.2^2}\)
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Phân tích từ số:
\(\frac{2^{12}.3^5-4^2.4^4.3^4}{2^{12}.3.3^5+4^2.4^4.3.3^4}=\frac{1}{6}\)
\(\frac{5^9.5.7^3-5^9.5.7^3.7}{5^9.7^3+5^9.2^3.7^3}=\frac{-10}{3}\)
Sau khi rút gọn là:
\(\frac{1}{6}-\left(-\frac{10}{3}\right)=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)
a) \(\left[-\frac{1}{2}\left(a-1\right)x^3y^4z^2\right]^5=\frac{-\left(a-1\right)^5}{32}x^{15}y^{20}z^{10}\)
Hệ số: \(\frac{-\left(a-1\right)^5}{32}\). Bậc của đơn thức: \(15+20+10=45\)
b) \(\left(a^5b^2xy^2z^{n-1}\right)\left(-b^3cx^4z^{7-n}\right)=-a^5b^5cx^5y^2z^6\)
Hệ số: \(-a^5b^5c\). Bậc của đơn thức: \(5+2+6=13\)
c) \(\left(-\frac{9}{10}a^3x^2y\right)\left(-\frac{5}{3}ax^5y^2z\right)^3=\left(-\frac{9}{10}a^3x^2y\right)\left(-\frac{125}{27}a^3x^{15}y^6z^3\right)\)\(=\frac{25}{6}a^6x^{17}y^7z^3\)
Hệ số: \(\frac{25}{6}a^6\). Bậc của đơn thức:\(17+7+3=27\)
Gợi ý : Phân tích hết ra thành tích các thừa số nguyên tố rồi đặt cái chung ra ngoài
-> rút gọn
-> kết quả
P/S : bài này cx ko dài lắm nhưg lười ^^
Ta có \(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.7^3.2^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2}{12}-\frac{5\left(-6\right)}{9}\)\(=\frac{1}{6}+\frac{30}{9}=\frac{7}{2}\)
Vậy A=7/2
A = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^6}\)
= \(\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\) = \(\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}\)= \(\frac{1}{6}-\frac{-10}{3}\)= 1/6 + 10/3 = 7/2
a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)
\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)
\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)
b,\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)
\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)
\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)
\(=3^n\cdot9+1-2^n\cdot4+1\)
\(=3^n\cdot10-2^n\cdot5\)
Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)
\(3^n\cdot10⋮10\)
Vậy : ....