[x+1]^2=16
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PT
0
BT
1
HP
1
17 tháng 12 2022
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+...+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
HH
0
VM
14 tháng 11 2019
b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)
c) \(\left(3x+2\right)^3=-27\)
\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+2=-3\)
\(\Rightarrow3x=\left(-3\right)-2\)
\(\Rightarrow3x=-5\)
\(\Rightarrow x=\left(-5\right):3\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!
30 tháng 8 2023
1: =(x+y-3x)(x+y+3x)
=(-2x+y)(4x+y)
2: =(3x-1-4)(3x-1+4)
=(3x+3)(3x-5)
=3(x+1)(3x-5)
3: =(2x)^2-(x^2+1)^2
=-[(x^2+1)^2-(2x)^2]
=-(x^2+1-2x)(x^2+1+2x)
=-(x-1)^2(x+1)^2
4: =(2x+1+x-1)(2x+1-x+1)
=3x(x+2)
5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(2x^2+2)*4x
=8x(x^2+1)
6: =(5x-5y)^2-(4x+4y)^2
=(5x-5y-4x-4y)(5x-5y+4x+4y)
=(x-9y)(9x-y)
7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)
=(x^2+2xy+y^2)(x^2-y^2)
=(x+y)^3*(x-y)
8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)
=[(x-2y)^2-4][(x+2y)^2-36]
=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)
KC
2
31 tháng 8 2023
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
31 tháng 8 2023
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
\(\left(x+1\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(\left(x+1\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)