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Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2=\frac{1}{2}\left[2\left(x+y\right)+\frac{1}{x}+\frac{1}{y}\right]^2\)
\(\Rightarrow P\ge\frac{1}{2}\left[2\left(x+y\right)+\frac{4}{x+y}\right]^2=18\)
\(\Rightarrow P_{min}=18\) khi \(x=y=\frac{1}{2}\)
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
\(M=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+1}\right)\)
\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2}{x^2+1}\)
b/ \(M=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
Do \(x^2+1\ge1\Rightarrow\frac{3}{x^2+1}\le3\Rightarrow1-\frac{3}{x^2+1}\ge1-3=-2\)
\(\Rightarrow M_{min}=-2\) khi \(x=0\)
Bài 5 :
a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
=> \(36x+3=0\)
=> \(x=-\frac{1}{12}\)
Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)
b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)
=> \(35x-5+60x-96+6x=0\)
=> \(101x-101=0\)
=> \(x=1\)
Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)
c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
=> \(\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(2x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)
=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)
=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)
=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)
=> \(-64x+123=0\)
=> \(x=\frac{123}{64}\)
Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)
\(A=\left(x-1\right)^4+\left(3-x\right)^4+6\left(x-1\right)^2\left(3-x\right)^2\)
Đặt \(\left\{{}\begin{matrix}x-1=a\\3-x=b\end{matrix}\right.\) \(\Rightarrow a+b=2\)
\(A=a^4+b^4+6a^2b^2=\left(a^2+b^2\right)^2+4a^2b^2\)
\(=\left[\left(a+b\right)^2-2ab\right]^2+4a^2b^2=\left[4-2ab\right]^2+4a^2b^2\)
\(=8a^2b^2-16ab+16=8\left(ab-1\right)^2+8\ge8\)
Dấu "=" xảy ra khi \(ab-1=0\Leftrightarrow\left(x-1\right)\left(3-x\right)-1=0\) \(\Rightarrow x=2\)
Hmmm... Mình biết làm rồi nhé :'> Nhưng các bạn có thể ghi cách làm của mình, cảm ơnnnnnnnnnnnn
\(A=x-2\sqrt{x-1}=\left(x-1\right)-2\sqrt{x-1}+1=\left(\sqrt{x-1}-1\right)^2\ge0\)
A min =0 khi \(\sqrt{x-1}-1=0\Leftrightarrow x-1=1\Leftrightarrow x=2\)
Vì \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0\)
Với \(x\ge0\) ta có:
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
\(\Rightarrow\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)
10 số x
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow1-\frac{1}{11}=11x-10x\)
\(\Rightarrow x=\frac{10}{11}\)
Vậy \(x=\frac{10}{11}\)
\(A=\left(\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}\right):\dfrac{x-1}{x^3}\)
\(=\dfrac{x^2+3}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)^2}\)
