Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Đây là câu a/
https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228
Còn câu b thì như sau:
Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)
\(A=\left[\dfrac{a^2-2a+4}{a-2}:\left(a^3+8\right)+\dfrac{a-2}{a^3+8}\cdot\dfrac{a^2-2a+4}{a^2-4}\right]\cdot\left(a^2-4\right)\)
\(=\left[\dfrac{a^2-2a+4}{a-2}\cdot\dfrac{1}{\left(a+2\right)\left(a^2-2a+4\right)}+\dfrac{a-2}{\left(a+2\right)\left(a^2-2a+4\right)}\cdot\dfrac{a^2-2a+4}{a^2-4}\right]\cdot\left(a^2-4\right)\)
\(=\left(\dfrac{1}{\left(a+2\right)\left(a-2\right)}+\dfrac{1}{\left(a+2\right)^2}\right)\cdot\left(a^2-4\right)\)
\(=\dfrac{a+2+a-2}{\left(a+2\right)^2\cdot\left(a-2\right)}\cdot\dfrac{\left(a+2\right)^2\cdot\left(a-2\right)^2}{1}\)
\(=2a\left(a-2\right)\)
Để A là số nguyên thì \(\left\{{}\begin{matrix}a\in Z\\a\notin\left\{2;-2\right\}\end{matrix}\right.\)
\(a^3+6=-3a-2a^2\)
\(\Rightarrow a^3+6+3a+2a^2=0\)
\(\Rightarrow a\left(a^2+3\right)+2\left(a^2+3\right)=0\)
\(\Rightarrow\left(a+2\right)\left(a^2+3\right)=0\)
Vì \(a^2+3>0\forall a\in R\) nên \(a+2=0\Leftrightarrow a=-2\)
\(A=\dfrac{a-1}{a+3}=\dfrac{-2-1}{-2+3}=\dfrac{-3}{1}=-3\)
Bài 1:
a^2-5ab-6b^2=0
=>a^2-6ab+ab-6b^2=0
=>a*(a-6b)+b(a-6b)=0
=>(a-6b)(a+b)=0
=>a=-b hoặc a=6b
TH1: a=-b
\(A=\dfrac{-2b-b}{-3b-b}+\dfrac{5b+b}{-3b+b}=\dfrac{-3}{-4}+\dfrac{6}{-2}=\dfrac{3}{4}-3=-\dfrac{9}{4}\)
TH2: a=6b
\(A=\dfrac{12b-b}{18b-b}+\dfrac{5b-6b}{18b+b}=\dfrac{11}{17}+\dfrac{-1}{19}=\dfrac{192}{323}\)
\(\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a^2-3a}\right):\dfrac{2a+2}{a}=\left(\dfrac{a-3}{a}-\dfrac{a}{a-3}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9}{a\left(a-3\right)}-\dfrac{a^2}{a\left(a-3\right)}+\dfrac{9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\left(\dfrac{a^2-6a+9-a^2+9}{a\left(a-3\right)}\right):\dfrac{2a+2}{a}=\dfrac{18-6a}{a\left(a-3\right)}:\dfrac{2a+2}{a}=\dfrac{6a-18}{\left(-a\right)\left(3-a\right)}:\dfrac{2a+2}{a}=\dfrac{6}{\left(-a\right)}:\dfrac{2a+2}{a}=\dfrac{6a}{\left(-2a^2\right)+\left(-2a\right)}.DKXD:a\ne0;a\ne3\)
thanks bạn