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dễ mà
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)thay x = 1;x=3 vào ra kq thui
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
b.\(A=\)viết lại đề nha bn
\(A=\frac{1^4+4}{3^4+4}.\frac{5^4+4}{7^4+4}...\frac{21^4+4}{23^4+4}\)
\(A=\frac{\left(4.1-3\right)^4+4}{\left(4.1-1\right)^4+4}.\frac{\left(4.2-3\right)^4+4}{\left(4.2-1\right)^4+4}...\frac{\left(4.6-3\right)^4+4}{\left(4.6-1\right)^4+4}\)
\(A=\frac{16.1^2-32.1+17}{16.1^2+1}.\frac{16.2^2-32.2+17}{16.2^2+1}....\frac{16.6^2-32.6+17}{16.6^2+1}\)
\(A=\frac{1}{17}.\frac{17}{65}.\frac{65}{145}....\frac{401}{577}=\frac{1}{577}\)
tíck mình nha bn thanks
a.\(a^4+a=a\left(a^3+1\right)=a\left(a+1\right)\left(a^2-a+1\right)\)
bai 1
a) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|-3,75=-2,,15\)
\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
Vậy ....
b) \(\left|\dfrac{5}{3}x\right|=\left|-\dfrac{1}{6}\right|\)
\(\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\-1\end{matrix}\right.\)
bai 2
a) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{1}{4}-\left|y\right|\)
\(\left|\dfrac{1}{6}+x\right|=\dfrac{1}{4}-\left|y\right|\) (*)
với mọi x ta luôn có \(\left|\dfrac{1}{6}+x\right|\ge0\)
\(\Rightarrow\dfrac{1}{4}-\left|y\right|\ge0\)
\(\Rightarrow\left|y\right|\le\dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{4}-\left|y\right|=\left|\dfrac{1}{4}-y\right|\)
Nên từ * \(\Rightarrow\left|\dfrac{1}{6}+x\right|=\left|\dfrac{1}{4}-y\right|\)
\(\Rightarrow\left|\dfrac{1}{6}+x\right|-\left|\dfrac{1}{4}-y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{6}+x=0\\\dfrac{1}{4}-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left|x-y\right|+\left|y+25\right|=0\)
với mọi x, y tao luôn có \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+25\right|\ge0\end{matrix}\right.\)
mà \(\left|x-y\right|+\left|y+25\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)
Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak
Bài 2:
\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)
\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)
\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)
\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)
\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)
\(=\dfrac{7.2.10}{81-4}\)
\(=\dfrac{720}{77}\)
a: \(\Leftrightarrow\dfrac{5}{2}:\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{15}{4}-3=\dfrac{3}{4}\)
\(\Leftrightarrow\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{5}{2}:\dfrac{3}{4}=\dfrac{5}{2}\cdot\dfrac{4}{3}=\dfrac{20}{6}=\dfrac{10}{3}\)
=>3/4x+1/2=10/3 hoặc 3/4x+1/2=-10/3
=>3/4x=17/6 hoặc 3/4x=-23/6
=>x=34/9 hoặc x=-46/9
b: \(\Leftrightarrow\dfrac{9}{4}:\left|x+\dfrac{1}{3}\right|=6.5-2=\dfrac{9}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{9}{4}:\dfrac{9}{2}=\dfrac{1}{2}\)
=>x+1/3=1/2 hoặc x+1/3=-1/2
=>x=1/6 hoặc x=-5/6
a)\(\frac{3}{x-4}-\frac{2}{4-x}=\frac{3}{x-4}+\frac{2}{x-4}=\frac{5}{x-4}\)
câu b làm tương tự nha bạn
c)\(\frac{3}{x+5}-\frac{2}{x+2}=\frac{3x+6-2x-10}{\left(x+5\right)\left(x+2\right)}=\frac{x-4}{\left(x+5\right)\left(x+2\right)}\)
d)\(\frac{9}{x-5}-\frac{6}{x^2-25}=\frac{9x+45-6}{x^2-25}=\frac{9x+39}{x^2-25}\)
mik làm hơi tắt bạn thông cảm nha
a) \(\dfrac{3}{x-4}-\dfrac{2}{4-x}\)
\(=\dfrac{3}{x-4}+\dfrac{2}{x-4}\)
\(=\dfrac{3+2}{x-4}\)
\(=\dfrac{5}{x-4}\)
b) \(\dfrac{7}{x-3}-\dfrac{4}{3-x}\)
\(=\dfrac{7}{x-3}+\dfrac{4}{x-3}\)
\(=\dfrac{7+4}{x-3}\)
\(=\dfrac{11}{x-3}\)
c) \(\dfrac{3}{x-5}-\dfrac{2}{x+2}\) MTC: \(\left(x-5\right)\left(x+2\right)\)
\(=\dfrac{3\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x+2\right)-2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x+6-2x+10}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{x+16}{\left(x-5\right)\left(x+2\right)}\)
d) \(\dfrac{9}{x-5}-\dfrac{6}{x^2-25}\)
\(=\dfrac{9}{x-5}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\) MTC: \(\left(x-5\right)\left(x+5\right)\)
\(=\dfrac{9\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9\left(x+5\right)-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+45-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+39}{\left(x-5\right)\left(x+5\right)}\)
a) Ta có: \(a^4+4=a^4+4a^2+4-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2=\left(a^2+2-2a\right)\left(a^2+2+2a\right)\)
ta thấy n4+4=(n2-2n+2)(n2+2n+2)=\(\left[\left(n-1\right)^2-1\right]\) \(\left[\left(n+1\right)^2+1\right]\)
Do đó B=\(\dfrac{\left(2^2+1\right)\left(4^2+1\right)}{\left(4^2+1\right)\left(6^2+1\right)}.\dfrac{\left(6^2+1\right)\left(8^2+1\right)}{\left(8^2+1\right)\left(10^2+1\right)}.....\dfrac{\left(18^2+1\right)\left(20^2+1\right)}{\left(20^2+1\right)\left(22^2+1\right)}=\dfrac{2^2+1}{22^2+1}=\dfrac{5}{485}=\dfrac{1}{97}\)