Tìm P biết
\(\dfrac{1}{2}P+\left(\dfrac{3}{4}\right)\left(4-2P\right)=-4\)
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28 tháng 2 2022
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
29 tháng 8 2017
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 ...
29 tháng 10 2021
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
1 tháng 10 2017
Ta có một số phân tích sau : \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)
Nhân mỗi biểu thức trong ngoặc ở cả tử thức với \(16\)\(=\)\(2^4\), ta được :
\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)
\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)
Kết hợp với phân tích nêu trên, khi đó :
\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)
\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Mình nhầm xíu =))
\(\dfrac{1}{2}P+\dfrac{3}{4}\left(4-2P\right)=-4\)
\(\Rightarrow\dfrac{1}{2}.\left[P+\dfrac{3}{2}\left(4-2P\right)\right]=-4\)
\(\Rightarrow P+6-3P=-4:\dfrac{1}{2}=-8\)
\(\Rightarrow6-2P=-8\)
=> 2P = 14
=> P = 7
\(\dfrac{1}{2}P+\dfrac{3}{4}.\left(4-2P\right)=4\)
\(\Rightarrow\dfrac{1}{2}.\left[P+\dfrac{3}{2}\left(4-2P\right)\right]=4\)
\(\Rightarrow P+6-3P=4:\dfrac{1}{2}=8\)
\(\Rightarrow6-2P=8\)
\(\Rightarrow2P=6-8=-2\)
=> P = -1