1/ 3x9 + 1/ 9x15 + 1/ 15x21 + 1/ 21x27 + ... + 1/39 x 45
Ai làm nhanh giúp mk vs mk đang gấp lắm
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=x+\dfrac{1}{x}\)
\(A=\left(\dfrac{x}{25}+\dfrac{1}{x}\right)+\dfrac{24}{25}x\ge2\sqrt{\dfrac{x}{25x}}+\dfrac{24}{25}.5=\dfrac{26}{5}\)
\(A_{min}=\dfrac{26}{5}\) khi \(x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
12x+3.23=23.x-4.32
12x+3.8=8.x-4.9
12x+24=8x-36
12x-8x=36-24
4x=12
x=12:4=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^x:1+2^x:2+...+2^x:49=2^{49}-1\)
\(2^x.1+2^x.\frac{1}{2}+...+2^x.\frac{1}{49}=2^{49}-1\)
\(2^x.\left(1+\frac{1}{2}+...+\frac{1}{49}\right)=2^{49}-1\)
Đặt: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{49}}\right)\)
=> \(A=1-\frac{1}{2^{49}}=\frac{2^{49}-1}{2^{49}}\)
\(2^{x-1}+2^{x-2}+2^{x-3}+...+2^{x-49}=2^{49}-1\)
<=> \(\frac{2^x}{2}+\frac{2^x}{2^2}+\frac{2^x}{2^3}+...+\frac{2^x}{2^{49}}=2^{49}-1\)
<=> \(2^x\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\right)=2^{49}-1\)
<=> \(2^x.\frac{2^{49}-1}{2^{49}}=2^{49}-1\)
<=> \(2^x=2^{49}\)
<=> x = 49.
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi S có n số hạng sao cho S = 1+ 2+ 3 + ...+ n = aaa ( a là chữ số)
\(\Rightarrow\)(n + 1).n : 2 = a.111
\(\Rightarrow\) n(n + 1) = a.222
\(\Rightarrow\) n(n + 1) = a.2.3.37
a là chữ số mà n; n + 1 là hai số tự nhiên liên tiếp nên a = 6
\(\Rightarrow\)n(n + 1) = 36.37 \(\Rightarrow\) n = 36
Vậy cần 36 số hạng.
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời :
Mk giúp bn câu a ) thôi mà sai thì thôi nhé :)))
a, \(\left|x\right|+\left|y\right|=0\)
\(\Leftrightarrow x=0;y=0\) \(\Rightarrow\left|x\right|+\left|y\right|=0\)
Vậy x = 0 ; y = 0
_Học tốt
câu a,b,c dạng tương tự nhau nha nên mình làm câu a
a)\(\left|x\right|+\left|y\right|=0\left(1\right)\)
Ta có: \(\hept{\begin{cases}\left|x\right|\ge0;\forall x,y\\\left|y\right|\ge0;\forall x,y\end{cases}\Rightarrow}\left|x\right|+\left|y\right|\ge0;\forall x,y\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}}\)
Vậy \(\left(x,y\right)=\left(0;0\right)\)
d) \(\left|x^2+1\right|=12\left(1\right)\)
Ta thấy \(x^2\ge0;\forall x\)
\(\Rightarrow x^2+1\ge1>0;\forall x\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x^2+1=12\)
\(\Leftrightarrow x^2=11\)
\(\Leftrightarrow x=\pm\sqrt{11}\)
Vậy \(x=\pm\sqrt{11}\)
\(\dfrac{1}{3\times9}+\dfrac{1}{9\times15}+\dfrac{1}{15\times21}+\dfrac{1}{21\times27}+...+\dfrac{1}{39\times45}\)
\(=\dfrac{1}{6}\times\left(\dfrac{6}{3\times9}+\dfrac{6}{9\times15}+\dfrac{6}{15\times21}+\dfrac{6}{21\times27}+...+\dfrac{6}{39\times45}\right)\)
\(=\dfrac{1}{6}\times\left(\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{27}+...+\dfrac{1}{39}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{6}\times\left(\dfrac{1}{3}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{6}\times\dfrac{14}{45}\)
\(=\dfrac{7}{135}\)
#AvoidMe
=1/6(6/3*9+6/9*15+...+6/39*45)
=1/6(1/3-1/9+1/9-1/15+...+1/39-1/45)
=1/6*14/45
=14/270=7/135