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Câu 1:
a: \(=\dfrac{1}{2}\cdot\dfrac{2n+1-2n+1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
b: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
1)5(x^2-1)+x(1-5x)= x-2
<=>5x2-5+x-5x2=x-2
<=>-5+x=x-2
<=>x-x=-2+5
<=>0x=3(vô lí)
vậy ko tìm được x
a: \(=24x^{2m-1+3-2m}y^{6-3m}-\dfrac{24}{7}y^{3n-7+6-3n}\cdot x^{3-2m}+8x^{3-2m+2m}\cdot y^{6-3n+3m}-24x^{3-2m}y^{6-2n+2}\)
\(=24x^2y^{6-3m}-\dfrac{24}{7}x^{3-2m}\cdot y^{-1}+8x^3y^{-3n+3m+6}-24x^{3-2m}y^{-2n+8}\)
b: \(=2x^{2n+1-2n}-6x^{2n+2-2n}+3x^{2n-1+1-2n}-9x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
a: \(=3x^{n+1}-2x^n-4x^n=3x^{n+1}-6x^n\)
b: \(=2x^{2n+1-2n}-6x^{2n+2-2n}+3x^{2n-1+1-2n}-9x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
\(\left(a-1\right)^{10}=\left(a-1\right)^{20}\)
\(\Rightarrow\left(a-1\right)^{20}-\left(a-1\right)^{10}=0\)
\(\Rightarrow\left(a-1\right)^{10}\cdot\left[\left(a-1\right)^{10}-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^{10}=0\\\left(a-1\right)^{10}-1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a-1=0\\\left(a-1\right)^{10}=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=1\\a-1=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a=1\\a=2\end{cases}}\)
a) Ta có:x-1=x-1
=>để \(\left(x-1\right)^{10}=\left(x-1\right)^{20}\)
thì x=1 hoặc x=2
vậy......
hc tốt