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Bài 2 :
a, \(\left|x-\frac{5}{3}\right|< \frac{1}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{3}< \frac{1}{3}\\x-\frac{5}{3}< -\frac{1}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 2\\x< \frac{4}{3}\end{cases}}}\)
b, \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\orbr{\begin{cases}\frac{2}{5}< x-\frac{7}{5}< \frac{3}{5}\\\frac{2}{5}< -x+\frac{7}{5}< \frac{3}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{9}{5}< x< 2\\1>x>\frac{4}{5}\end{cases}}\)
a) \(\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Rightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Rightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0^{10}\\\left(x-5\right)^2=0+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+5\\\left(x-5\right)^2=1^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=1+5\\x=-1+5\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=4\\x=6\end{cases}}\)
Vậy x = 4 hoặc x = 5 hoặc x = 6
\(a)\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Leftrightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Leftrightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-4\right)\left(x-6\right)=0\end{cases}}\)
[ ra \(\left(x-4\right)\left(x-6\right)\)do \(\left(x-5\right)^2-1=\left(x-5-1\right)\left(x-5+1\right)=\left(x-6\right)\left(x-4\right)\)]
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=6\end{cases}}\)
_Minh ngụy_
c) -12 . ( x - 5 ) + 7 . ( 3 -x ) = 5
<=> -12.x + 60 + 21 -7.x = 5
<=> -19 .x + 81 = 5
<=> -19.x = 5 - 81
<=> -19.x = -76
<=> x = -76 : -19
<=> x = 4
Vậy x = 4
d) 30(x+2)-6(x-5)-24x=100
<=> 30.x + 60 - 6.x + 30 -24.x = 100
<=> 0 + 90 = 100
<=> 90 = 100
<=> x \(\in\varnothing\)
Vậy x \(\in\varnothing\)
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
a) 3x = 33.35 = 38
=> x = 8
b) 2x.2 = 222
2x+1 = 22
=> x + 1 = 22
x = 21
c) (7x-11)3 = 25.52 + 200 = 800 + 200 = 1 000 = 103
=> 7x -11 = 10
7x = 21
x = 3
d) 6x = 362 = (62)2 = 64
=> x = 4
e) 64.4x = 49
4x = 49/64
f) (21-1)3 = 203 ( xem lại đề)
h) 64.4x.2 = 45
4x.128 = 45
4x = 45/128
a) \(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)
\(\Rightarrow\left(x-5\right)=\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\left(x-5\right)=\frac{11}{15}\)
\(\Rightarrow x-5=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}+5\)
\(\Rightarrow x=\frac{86}{15}\)
b) \(\frac{2}{3}\cdot x-\frac{3}{2}\cdot x=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\left(-\frac{5}{6}\right)\)
\(\Rightarrow x=-\frac{1}{2}\)
c) \(-\frac{2}{3}\cdot x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{3}{20}\)
d) \(4-\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=-\frac{1}{5}\)
\(\Rightarrow\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=4-\left(-\frac{1}{5}\right)\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x+\frac{3}{4}=\frac{21}{5}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{21}{5}-\frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{69}{20}\)
\(\Rightarrow\)\(x=\frac{69}{20}:\frac{1}{2}\)
\(\Rightarrow\)\(x=\frac{69}{10}\)
6/7+5/8÷5-3/16×(-2)²
=6/7+1/8-3/4
=55/56-3/4
=13/56
b.2/3 + 1/3.( -4/9 + 5/6 ) : 7/12
=2/3 + 1/3. ( -8/18 + 15/18 ) : 7/12
=2/3 + 1/3 . 7/18 : 7/12
=2/3 + 7/54 : 7/12
= 2/3 + 2/9
=6/9 + 2/9
= 8/9
\(a,\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot(-2)^2\)
\(=\frac{6}{7}+\frac{5}{8}:\frac{5}{1}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot4}{16}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot1}{4}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{48+7-42}{56}=\frac{13}{56}\)
\(b,\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-2}{3}+\frac{5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-4+5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}:\frac{2}{3}=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}\cdot\frac{3}{2}=\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
c, Xem lại đề
d, \(\frac{-3}{5}+\left[\frac{-2}{5}-99\right]\)
\(=\frac{-3}{5}+\frac{-497}{5}=\frac{-500}{5}=-100\)
b, Tìm x
\(\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{1}{7}-\frac{1}{8}\right]\cdot56\)
\(\Rightarrow\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{8}{56}-\frac{7}{56}\right]\cdot56\)
\(\Rightarrow\left[\frac{6}{33}+\frac{11}{33}\right]\cdot x=1\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=1\cdot\frac{33}{17}=\frac{33}{17}\)
Sửa đề P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)CM P < 5/16
=> 5P = \(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\)
Lấy 5P trừ P theo vế ta có
5P - P = \(\left(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\right)\)
4P = \(1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{11}{5^{10}}-\frac{10}{5^{10}}\right)-\frac{11}{5^{11}}\)
4P = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
Đặt Q = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
=> 5Q \(=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
Lấy 5Q trừ Q theo vế ta có
5Q - Q = \(\left(5+1+\frac{1}{5}+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)\)
4Q \(=5-\frac{1}{5^{10}}\)
=> Q\(=\frac{5}{4}-\frac{1}{5^{10}.4}\)
Khi đó 4P = \(\frac{5}{4}-\frac{1}{5^{10}.4}-\frac{11}{5^{11}}\)
=> P = \(\frac{5}{16}-\frac{1}{5^{10}.16}-\frac{11}{5^{11}.4}\)
\(=\frac{5}{16}-\frac{1}{5^{10}}\left(\frac{1}{16}-\frac{11}{5.4}\right)< \)\(\frac{5}{16}\)
Bài làm:
Ta có: \(\frac{1}{5^2}+\frac{2}{5^2}+\frac{3}{5^2}+...+\frac{11}{5^2}\)
\(=\frac{1+2+3+...+11}{5^2}=\frac{\left(1+11\right).11:2}{5^2}=\frac{66}{25}>1>\frac{1}{16}\)
\(\Rightarrow P>\frac{1}{16}\)
=> Đề sai