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\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
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2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
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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.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
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Áp dụng bđt Schwarz ta có: \(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\dfrac{a+b+c}{2}=1\).
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Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
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Từ giả thiết:
\(a+b+1=8ab\le2\left(a+b\right)^2\)
\(\Rightarrow2\left(a+b\right)^2-\left(a+b\right)-1\le0\)
\(\Rightarrow\left(a+b-1\right)\left(2a+2b+1\right)\le0\)
\(\Rightarrow a+b-1\le0\) (do \(2a+2b+1>0\))
\(\Rightarrow1\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le\dfrac{1}{4}\Rightarrow\dfrac{1}{ab}\ge4\)
Ta có:
\(A=\dfrac{a^2+b^2}{a^2b^2}\ge\dfrac{2ab}{a^2b^2}=\dfrac{2}{ab}\ge2.4=8\)
\(A_{min}=8\) khi \(a=b=\dfrac{1}{2}\)
cho em hỏi là \(a+b+1=8ab\) ≤ \(2\left(a+b\right)^2\)
vì sao ạ? em chưa có hiểu lắm
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c/ Ta có:\(6a-5b=1\)
\(\Rightarrow5b=6a-1\)
Theo đề thì: \(A=4a^2+\left(6a-1\right)^2=40a^2-12a+1\)
\(=\left(\left(2\sqrt{10}a\right)^2-\frac{2.2.\sqrt{10}.3a}{\sqrt{10}}+\frac{9}{10}\right)+\frac{1}{10}\)
\(=\left(2\sqrt{10}a-\frac{3}{\sqrt{10}}\right)^2+\frac{1}{10}\ge\frac{1}{10}\)