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a) \(\left|2x-1\right|=5\)
\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)
b) \(\left(5^x-1\right)3-2=70\)
\(\Rightarrow5^x.3-3=72\)
\(\Rightarrow5^x.3=75\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
c) \(\left(x-1\frac{1}{2}\right)^2+\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)
............. Làm tiếp nhé!
d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
a, \(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=0+\frac{1}{16}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2=\left(\frac{-1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{3}=\frac{-1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{11}{12}\\\frac{1}{x}=\frac{5}{12}\end{cases}\Rightarrow\orbr{\begin{cases}11x=12\\5x=12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}}\)
b, \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
Đặt S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)
2S = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}\)
2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2S = \(\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
S = \(\frac{22}{45}:2=\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}\Rightarrow x=\frac{23}{11}\)
a/ (1/x -2/3)2=1/16=(1/4)2
Có 2 trường hợp:
+/ 1/x -2/3= - 1/4
<=> 1/x =2/3 -1/4 = 5/12
=> x1=12/5
+/ 1/x - 2/3 =1/4
<=> 1/x = 2/3 +1/4= 11/12
=> x2=12/11
b/ Ta có:
2/(1.2.3)=1/(1.2) - 1/2.3 ; 2/(2.3.4)=1/2.3 -1/3.4 ; ...; 2/(8.9.10)=1/8.9 -1/9.10
=> (1/1.2.3 + 1/2.3.4 +...+1/8.9.10)=23/45
<=> (1/1.2 -1/2.3 +1/2.3 -1/3.4 +...+1/8.9-1/9.10).x/2=23/45
<=> (1/1.2 -1/9.10).x/2 =23/45
<=> x.11/45=23/45
=> x=23/11
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)
=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)
=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)
=>24<28x<231
=>28x\(\in\){25;26;27;28;.............................;230}
=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224
=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}
Vậy x\(\in\) {1;2;3;4;5;6;7;8}
\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)
\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)
\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)
=>3\(⋮\)\(\frac{1}{6}+x\)
=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}
Ta có bảng:
| \(\frac{1}{6}+x\) | -1 | 1 | -3 | 3 |
| x | \(-1\frac{1}{6}\) | \(1\frac{1}{6}\) | \(-3\frac{1}{6}\) | 3\(\frac{1}{6}\) |
Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}
Chúc bn học tốt
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
Các bạn ơi câu 2 mik chép đề sai nha:
đề đúng đây
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)