P= (1-1/2).(1-1/3).(1-1/4)...(1-1/50) Giúp mik nhanh vs ạ
Tính : C=1/2^2+1/2^4+1/2^6+1/2^8+…+1/2^50
Giúp mik vs ạ, cần gấp
B= 1/2 + 1/3 - 1/4 +...- 1/2022 + 1/2023 C= 1/1012 + 1/1013+...+ 1/2022 + 1/2023
Tính: B-C
mọi người sửa nhanh giúp mik vs ạ
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
2 . ( x³ -1)-2x²(x+2x⁴) +(4x⁵+4)x=6. 3. (X²-4x+16)(x+4)-x(x+1)(x+2)+3x²= 0 4 . ( 8x +2 ) (1-3x) + ( 6x-1)(4x-10) =-50 Đề bài là tìm x nha mn Nhanh giúp mik vs
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
Tính nhanh:(\(1-\dfrac{1}{2^2}\)).(1-\(\dfrac{1}{3^2}\)).(1-\(\dfrac{1}{4^2}\))...(1-\(\dfrac{1}{10^2}\))
mik đag cần rất gấp giúp mik vs
Giải:
(1-1/22).(1-1/32).(1-1/42).....(1-1/102)
=3/2.2 . 8/3.3 . 15/4.4 . ... . 99/10.10
=1.3.2.4.3.5.....9.11/2.2.3.3.4.4.....10.10
=1.2.3.....9/2.3.4.....10 . 3.4.5....11/2.3.4.....10
=1/10.11/2
=11/20
Chúc bạn học tốt!
1/2a^2 +1 :x = 2 Giúp nhanh hộ mik vs ạ
\(8(x+\dfrac{1}{x} )^{2} \)\(+4(x^{2}+\dfrac{1}{x^{2} } )^{2}\)\(-4 (x^{2}+\dfrac{1}{x^{2}} )(x+\dfrac{1}{x})^{2} \)\(=(x+4)^{2}\)
giúp mik vs ạ cho mik cách giải pt này vs ạ
=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2
Đặt x+1/x=a(a>=2)
=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2
=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2
=>(x+4)^2=16
=>x+4=4 hoặc x+4=-4
=>x=-8;x=0
Điều kiện: \(x\ne0\)
\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)
a) 1/4 - 3 (1/12 + 3/8 )
b) ( -2/3 + 3/5 ) : 1/50 - 30
mọi người giải giúp em vs ạ em cần gấp ạ. xin cảm ơn
a) \(\dfrac{1}{4}-3\left(\dfrac{1}{12}+\dfrac{3}{8}\right)=\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{9}{8}=-\dfrac{9}{8}\)
b) \(\left(-\dfrac{2}{3}+\dfrac{3}{5}\right):\dfrac{1}{50}-30=\left(-\dfrac{2}{3}+\dfrac{3}{5}\right).50-30=-\dfrac{100}{3}+30-30=-\dfrac{100}{3}\)
tìm x biết: 3.|4x-1|-2=19 giúp mik nhanh vs ạ
\(3\left|4x-1\right|-2=19\)
\(3\left|4x-1\right|=21\)
\(\left|4x-1\right|=7\)
⇔\(\left[{}\begin{matrix}4x-1=7\\4x-1=-7\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left|4x-1\right|=21:3=7\\ \Rightarrow\left[{}\begin{matrix}4x-1=7\\4x-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(3.\left|4x-1\right|-2=19\)
<=> \(\left[{}\begin{matrix}3.\left(4x-1\right)-2=19\left(x\ge\dfrac{1}{4}\right)\\3.\left[-\left(4x-1\right)\right]-2=19\left(x< \dfrac{1}{4}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}12x-3-2=19\\3.\left(-4x+1\right)-2=19\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\left(TM\right)\\x=-\dfrac{3}{2}\left(TM\right)\end{matrix}\right.\)