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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 ...
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 \(\left|x-y\right|+\left|y+25\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)
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)\(\frac{3}{x-4}-\frac{2}{4-x}=\frac{3}{x-4}+\frac{2}{x-4}=\frac{5}{x-4}\)
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: \(\Leftrightarrow\dfrac{5}{2}:\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{15}{4}-3=\dfrac{3}{4}\)
\(\Leftrightarrow\left|\dfrac{3}{4}x+\dfrac{1}{2}\right|=\dfrac{5}{2}:\dfrac{3}{4}=\dfrac{5}{2}\cdot\dfrac{4}{3}=\dfrac{20}{6}=\dfrac{10}{3}\)
=>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
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
a) −x2+21−5x−x2+21−5x =x2+2−(1−5x)=x2+2−(1−5x) =x2+25x−1=x2+25x−1;
b) −4x+15−x−4x+15−x =4x+1−(5−x)=4x+1−(5−x) =4x+1x−5
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=\(\dfrac{\left(2^2+1\right)\left(4^2+1\right)}{\left(4^2+1\right)\left(6^2+1\right)}.\dfrac{\left(6^2+1\right)\left(8^2+1\right)}{\left(8^2+1\right)\left(10^2+1\right)}.....\dfrac{\left(18^2+1\right)\left(20^2+1\right)}{\left(20^2+1\right)\left(22^2+1\right)}=\dfrac{2^2+1}{22^2+1}=\dfrac{5}{485}=\dfrac{1}{97}\)