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\(=\left[\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right)+\frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right)+\frac{1}{5}\left(\frac{1}{14}-\frac{1}{19}\right)+...+\frac{1}{5}\left(\frac{1}{44}-\frac{1}{49}\right)\right]\cdot\frac{1-\left(3+5+...+49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(52+52+...+52\right)\left\{12\text{ số 52}\right\}}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)
\(=\frac{9}{196}\cdot-7=\frac{9}{28}\)
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
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) (x + 2)2 = 81
=> (x + 2)2 = 92
=> \(\orbr{\begin{cases}x+2=-9\\x+2=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\x=7\end{cases}}\)
b) 5x + 5x + 2 = 650
=> 5x + 5x . 52 = 650
=> 5x + 5x . 25 = 650
=> 5x (25 + 1) = 650
=> 5x . 26 = 650
=> 5x = 650 : 26
=> 5x = 25
=> 5x = 52
=> x = 2
d) (2x - 1)2 - 5 = 20
=> (2x - 1)2 = 25
=> (2x - 1)2 = 52
=> \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
g) (x - 1)3 = (x - 1)
=> (x - 1)3 - (x - 1) = 0
=> (x - 1) .[(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 1
=> x = 2
Nếu x - 1 = -1
=> x = 0
Vậy \(x\in\left\{0;1;2\right\}\)
(\(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)).\(\frac{1-3-5-...-49}{89}\)
= \(\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{45.49}\right).\frac{1-3-5-...-49}{89}\)
\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\frac{24.\left(49+3\right)}{2}}{89}\)
\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\left(-7\right)\)
\(=-\frac{9}{28}\)
Có chỗ ghi nhầm 44 thành 45. Tự sửa nhé
Bài 2/ a/
|2x + 3| = x + 2
Điều kiện \(x\ge-2\)
Với x < - 1,5 thì ta có
- 2x - 3 = x + 2
<=> 3x = - 5
<=> \(x=-\frac{5}{3}\)
Với \(x\ge-1,5\)thì ta có
2x + 3 = x + 2
<=> x = - 1
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(49+3\right)\left(\left(49-3\right):2+1\right):2}{89}\)
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-26.24}{89}=\frac{45}{4.5.49}.\frac{-623}{89}=-\frac{9}{28}\)
\(A=\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(\Rightarrow5A=5.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(=\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right).\frac{1+\frac{\left(-3-47\right).23}{2}-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1+\left(-575\right)-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}=\frac{45}{196}.\left(-7\right)=-\frac{45}{26}\)
Bài 1:
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right).-\left(\frac{3+5+7+...+49-1}{89}\right)\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).-\left(\frac{\left(49+3\right).24:2-1}{89}\right)\)(Do tổng có 24 số)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).-\left(\frac{52.12-1}{89}\right)\)
\(=\frac{1}{5}.\frac{45}{196}.\left(-7\right)=-\frac{9}{28}\)
Bài 2:
a) Ta có:
\(|2x+3|=x+2\)
<=> x + 2 >=0 và: \(\orbr{\begin{cases}2x+3=x+2\\2x+3=-x-2\end{cases}}\)
<=> x >= -2 và \(\orbr{\begin{cases}2x-x=2-3\\2x+x=-2-3\end{cases}}\)
<=> x >= -2 và \(\orbr{\begin{cases}x=-1\left(n\right)\\x=-\frac{5}{3}\left(n\right)\end{cases}}\)( n là viết tắt của "nhận" nha bạn)
Vậy x ={-1 ; -5/3}
Xin lỗi vì tớ ko thể lồng dấu \(\hept{\begin{cases}\\\end{cases}}\) và dấu \(\orbr{\begin{cases}\\\end{cases}}\) được nếu lồng sẽ bị lỗi nên tớ dùng chữ "và" nha bạn
b)
A = \(|x-2006|+|2007-x|\)
Vì \(\hept{\begin{cases}|x-2006|\ge0\\|2007-x|\ge0\end{cases}}\)
Nến giá trị A sẽ nhỏ nhất khi \(\orbr{\begin{cases}x=2006\\x=2007\end{cases}}\)
=> Min A = 1 khi x ={2006 ; 2007}