\(A=\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}+2\right)}{\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)}+\dfrac{1}{2-\sqrt{a}}\left(ĐK:a\ge0;a\ne4\right)\\ =\dfrac{\sqrt{a}-3}{2-\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}\\ =\dfrac{\sqrt{a}-2}{-\left(\sqrt{a}-2\right)}=-1\)

Vậy `A=-1`

\(\left(x^2-6x+9\right)+15\left(x^2-6x+10\right)=1\)

\(\Leftrightarrow\left(x-3\right)^2+15\left[\left(x-3\right)^2+1\right]=1\)

\(\Leftrightarrow16\left(x-3\right)^2+15=1\)

\(\Leftrightarrow16\left(x-3\right)^2=-14\)

=> Phương trình vô nghiệm 

\(\left(x^2-6x+9\right)-15\left(x^2-6x+10\right)=1\)

Đặt : \(x^2-6x+9=\left(x-3\right)^2=t\) thay vào pt ta được :

\(t^2-15\left(t+1\right)=1\)

\(\Leftrightarrow t^2-15t-16=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-16\right)=0\)

\(\Leftrightarrow t=\left\{{}\begin{matrix}16\\-1\end{matrix}\right.\)

với : \(t=-1\) thì \(\left(x-3\right)^2=-1\)

\(\Rightarrow ptvonghiem\)

Với : \(t=16\) thì \(\left(x-3\right)^2=16\)

\(\Leftrightarrow x\in\left\{{}\begin{matrix}7\\-1\end{matrix}\right.\)

\(vay...\)

\(Từ:gt\) \(a+b+c=0\)

\(\Rightarrow b+c=-a\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

cmt tương tự với :

\(b^2-a^2-c^2=2ac\)

\(c^2-a^2-b^2=2ab\)

\(\Rightarrow A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

\(\Rightarrow A=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)

\(\Rightarrow A=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)

\(\Rightarrow A=\dfrac{3abc}{2abc}\)

\(\Rightarrow A=\dfrac{3}{2}\)

Đặt : \(a=1-\dfrac{2x-1}{x+1},ta\)  có :

\(\dfrac{12\left(2x-1\right)}{x+1}-20=-12\left(1-\dfrac{2x-1}{x+1}\right)-8=-12a-8\)

Do đó pt có dạng : \(a^3+6a^2=-12a-8\)

\(\Leftrightarrow\left(a-2\right)^3=0\)

\(\Leftrightarrow a=-2\)

vậy pt đã cho tương đương với :

\(1-\dfrac{2x-1}{x+1}=2\) hay \(\dfrac{2x-1}{x+1}=3\)

\(\Leftrightarrow2x-1=3\left(x+1\right)\)

\(\Leftrightarrow x=-4\)

Vậy ...

Đặt (2x-1)/(x+1) = a, để rút gọn vế giải cho dễ

(1-a)^3 + 6a = 12a - 20

Tìm a, từ đó giải ra x

\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)

\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)

mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)

Khử mẫu ta được :

\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)

do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)

\(Vậy...\)

\(Ta\) \(có:\) \(1+a^2=ab+bc+ca+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(c+a\right)\)

\(1+b^2=ab+bc+ca+b^2=\left(a+b\right)\left(b+c\right)\)

\(1+c^2=ab+bc+ca+c^2=\left(a+c\right)\left(c+b\right)\)

\(Khi\) \(đó:\) \(A=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}\)

\(\Rightarrow A=1\)

\(\left(x-2\right)\left(x-1\right)\left(x-4\right)\left(x-8\right)=4x^2\)

\(\Leftrightarrow[\left(x-2\right)\left(x-4\right)][\left(x-1\right)\left(x-8\right)]=4x^2\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-9x+8\right)=4x^2\)

thấy \(x=0;2\) không phải nghiệm của phương trình nên ta chia hai vế của pt cho \(x^2\) ta được \(:\)

\(\Leftrightarrow\left(x+\dfrac{8}{x}-9\right)\left(x+\dfrac{8}{x}-6\right)=4\)

\(Đặt:\) \(x+\dfrac{8}{x}=a\) thì pt trở thành \(:\)

\(\left(a-6\right)\left(a-9\right)=4\)

\(\Leftrightarrow a^2-15a+50=0\)

\(\Leftrightarrow\left(a-5\right)\left(a-10\right)=0\Leftrightarrow\left\{{}\begin{matrix}a=5\\a=10\end{matrix}\right.\)

\(Với\) \(a=5\) thì \(x+\dfrac{8}{x}=5\Leftrightarrow x^2-5x+8=0\left(vônghiem\right)\)

\(Với\) \(a=10\) thì \(x+\dfrac{8}{x}=10\Leftrightarrow x^2-10x+8=0\Leftrightarrow\left\{{}\begin{matrix}x=5-căn17\\x=5+căn17\end{matrix}\right.\)

\(Vậy...\)

căn bậc 2 của \(17\) đấy á

Lời giải:
ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow x^2-4x+21-6\sqrt{2x+3}=0$

$\Leftrightarrow (x^2-6x+9)+[(2x+3)-6\sqrt{2x+3}+9]=0$

$\Leftrightarrow (x-3)^2+(\sqrt{2x+3}-3)^2=0$

Ta thấy: $(x-3)^2\geq 0; (\sqrt{2x+3}-3)^2\geq 0$ với mọi $x\geq \frac{-3}{2}$

Do đó để tổng của chúng bằng $0$ thì:
$(x-3)^2=(\sqrt{2x+3}-3)^2=0$

$\Leftrightarrow x=3$ (tm)

Theo đề ra, ta có:

\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Leftrightarrow\left(a^{100}+b^{100}\right).\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)+a^{202}+b^{202}=a^{202}+b^{202}+2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2\right)=2a^{101}.b^{101}\)

\(\Leftrightarrow a^{100}.b^{100}.\left(a^2+b^2-2ab\right)=0\)

\(\Leftrightarrow a=b=0\)

\(\Rightarrow a^{100}+b^{100}=a^{101}+b^{101}\)

\(\Rightarrow a^{100}=a^{101}\)

\(\Leftrightarrow a^{100}.\left(a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow A=a^{2015}+b^{2015}=1+1=2\).

\(Từ:\) \(a^{100}+b^{100}=a^{101}+b^{101}\)

\(\Leftrightarrow a^{100}\left(a-1\right)+b^{100}\left(b-1\right)=0\left(1\right)\)

\(và\) \(a^{101}+b^{101}=a^{102}+b^{102}\)

\(\Leftrightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)=0 \left(2\right)\)

\(Từ\left(1\right)\) \(và\) \(\left(2\right)\)

\(\Rightarrow a^{101}\left(a-1\right)+b^{101}\left(b-1\right)-a^{100}\left(a-1\right)-b^{100}\left(b-1\right)=0\)

\(\Leftrightarrow a^{100}\left(a-1\right)^2+b^{100}\left(b-1\right)^2\)

\(Do\) \(a,b>0\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

\(\Rightarrow A=1+1=2\)

em không chắc cho lắm ạ

ĐKXĐ : a;b;c  \(\ne0\)

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2000}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}-\dfrac{1}{a}\)

\(\Leftrightarrow\dfrac{b+c}{bc}=\dfrac{-\left(b+c\right)}{a\left(a+b+c\right)}\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{1}{bc}+\dfrac{1}{a\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a\left(a+b+c\right)+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a^2+ab+ac+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\dfrac{\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\a+c=0\end{matrix}\right.\left(1\right)\)

Từ (1) kết hợp a + b + c = 2000 ta được điều phải chứng minh