tính : \(\left(x^2-4\right)^2-\left(x-2\right).\left(x+2\right).\left(x^2+4\right)\)
K
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22 tháng 12 2018
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
25 tháng 5 2022
a)
\(\begin{matrix}N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\^-M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\\overline{N\left(x\right)-M\left(x\right)=-3x^4+18x^3-2x^2-4x-1}\end{matrix}\)
b)
\(\begin{matrix}M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\^+N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\\overline{M\left(x\right)+N\left(x\right)=-5x^4+14x+\dfrac{5}{3}}\end{matrix}\)
AH
Akai Haruma
Giáo viên
16 tháng 9 2023
Lời giải:
Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$
Đề bài
$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow 12=x$ (thỏa mãn)
8 tháng 1 2022
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
8 tháng 1 2022
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 10 2021
\(5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2=\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]\)
\(\left(y-x\right)^2=\left(x-y\right)^2\)
\(\Rightarrow\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]:\left(y-x\right)^2=5\left(x-y\right)^2-3\left(x-y\right)+4\)
\(\left(x^2-4\right)^2-\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ =\left(x^2-4\right)^2-\left(x^2-4\right)\left(x^2+4\right)\\ =\left(x^2-4\right)\left(x^2-4-x^2-4\right)\\ =-8\left(x^2-4\right)\)