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\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
\(\left(\dfrac{2a^3+a^2-a}{a^3-1}-2+\dfrac{1}{1-a}\right):\left(1:\dfrac{2a-1}{a-a^2}\right)\)
\(=\left(\dfrac{2a^3+a^2-a-2a^3+2-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}\right):\left(\dfrac{a\left(1-a\right)}{2a-1}\right)\)
\(=\dfrac{-2a+1}{\left(a-1\right)\left(a^2+a+1\right)}.\dfrac{2a-1}{a\left(1-a\right)}\)
\(=\dfrac{6a-3}{\left(a-1\right)^2\left(a^2+a+1\right)}\)
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Thấy sai sai :vv
a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
\(C=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{a+1}{a}\right):\dfrac{a-1}{a^3}\)
\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right):\dfrac{a-1}{a^3}\)
\(=\dfrac{a+2\cdot\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)
\(=\dfrac{2a\left(a+1\right)}{\left(a^2+1\right)\cdot\left(a+1\right)^3}\cdot\dfrac{a^2}{a-1}\)
\(=\dfrac{2a^3}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)
Đặt: \(L=\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}+\dfrac{2a^2-a-10}{a^3-a^2+a-1}\)
Ta có:
\(\dfrac{3\left(a+2\right)}{a^3+a^2+a+1}=\dfrac{3\left(a+2\right)}{a^2\left(a+1\right)+1\left(a+1\right)}=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}\)
\(\dfrac{2a^2-a-10}{a^3-a^2+a-1}=\dfrac{a\left(2a-1\right)-10}{a^2\left(a-1\right)+1\left(a-1\right)}=\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Như vậy \(L=\dfrac{3\left(a+2\right)}{\left(a^2+1\right)\left(a+1\right)}+\dfrac{a\left(2a-1\right)-10}{\left(a^2+1\right)\left(a-1\right)}\)
Đặt:
\(N=\dfrac{5}{a^2+1}+\dfrac{3}{2a+2}-\dfrac{3}{2a-2}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{3\left(2a-2\right)}{\left(2a+2\right)\left(2a-2\right)}-\dfrac{3\left(2a+2\right)}{\left(2a+2\right)\left(2a-2\right)}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6}{4a^2-4}-\dfrac{6a+6}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{6a-6-6a-6}{4a^2-4}=\dfrac{5}{a^2+1}+\dfrac{-12}{4a^2-4}\)
\(N=\dfrac{5}{a^2+1}+\dfrac{-12}{4\left(a^2-1\right)}=\dfrac{5}{a^2+1}+\dfrac{-3}{a^2-1}\)
\(N=\dfrac{5\left(a^2-1\right)}{\left(a^2+1\right)\left(a^2-1\right)}+\dfrac{-3\left(a^2+1\right)}{\left(a^2-1\right)\left(a^2+1\right)}\)
\(N=\dfrac{5a^2-5-3a^2-3}{a^4-1}=\dfrac{2a^2-8}{a^4-1}\)
Thay M với N vào A Mình cạn sức rồi
Cảm ơn nhiều!!!!