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1: =>3x+1=4
=>3x=3
hay x=1
2: \(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^9}{98^3}=\dfrac{1}{2^3}+\dfrac{7^9}{7^6\cdot2^3}\)
\(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^3}{2^3}=\dfrac{344}{2^3}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
3: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\x-\dfrac{2}{9}=-\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{9}\end{matrix}\right.\)
4: =>x+2=0 và y-1/10=0
=>x=-2 và y=1/10
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
...
\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)
Vậy ...
b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
...
\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)
Vậy ...
áp dụng hằng đẳng thức (a+b)(a-b)=a^2-b^2 Minh Hoang Hai
a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)
ĐKXĐ : \(x\ne1;x\ne2\)
(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow4x+7=15\\\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)
Vậy pt vô nghiệm .
b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )
ĐKXĐ : \(x\ne3;x\ne-2\)
(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={0}.
c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)
ĐKXĐ : \(x\ne1;x\ne3\)
\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow18-6x+4x-4=8\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy tập nghiệm của phương trình là S={-3}
d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)
ĐKXĐ : \(x\ne0;x\ne2\)
\(\left(4\right)\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={-1}
a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )
\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)
\(\Leftrightarrow\) \(2-x+5x+5=15\)
\(\Leftrightarrow\)\(4x=8\)
\(\Rightarrow\) \(x=2\) ( KTM )
S = ∅
b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )
\(\Leftrightarrow\) \(\left(x+2\right)\left(3-x\right)+x\left(x+2\right)=5x+2\left(3-x\right)\)
\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)
\(\Leftrightarrow\) \(3x+6=3x+6\)
\(\Rightarrow\)\(0x=0\) ( TM )
\(\Rightarrow\) Phương trình vô số nghiệm
S = R
c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )
\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow\) \(18-6x+4x-4=8\)
\(\Leftrightarrow\) \(-2x=-6\)
\(\Rightarrow x=3\) ( KTM )
S = ∅
d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )
\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)
\(\Leftrightarrow\) \(x^2+2x-x+2=2\)
\(\Leftrightarrow\) \(x^2+x=0\)
\(\Leftrightarrow\) \(x\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
S = \(\left\{2\right\}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak
Bài 2:
\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)
\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)
\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)
\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)
\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)
\(=\dfrac{7.2.10}{81-4}\)
\(=\dfrac{720}{77}\)
Thừa số tổng quát:
\(1+\dfrac{1}{n^2+2n}=\dfrac{n^2+2n+1}{n^2+2n}=\dfrac{\left(n+1\right)^2}{\left(n+1\right)^2-1}\)
Đặt: \(\left(n+1\right)^2=t\ge0\) biểu thức được phát biểu dưới dạng: \(\dfrac{t}{t-1}\) Thay vào bài toán tìm được giá trị.