Theo giả thiết, ta có: \(a^2b^2+b^2c^2+c^2a^2=a^2b^2c^2\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\)

Áp dụng BĐT AM - GM cho 5 số, ta được: \(\hept{\begin{cases}a.a.a.b.b\le\frac{a^5+a^5+a^5+b^5+b^5}{5}=\frac{3a^5+2b^5}{5}\\b.b.b.a.a\le\frac{b^5+b^5+b^5+a^5+a^5}{5}=\frac{3b^5+2a^5}{5}\end{cases}}\)

\(\Rightarrow\frac{5\left(a^5+b^5\right)}{5}\ge a^2b^2\left(a+b\right)\)hay \(a^5+b^5\ge a^2b^2\left(a+b\right)\)

\(\Rightarrow\frac{1}{\sqrt{a^5+b^5}}\le\frac{1}{ab\sqrt{a+b}}\)(1) .

Tương tự, ta có: \(\frac{1}{\sqrt{b^5+c^5}}\le\frac{1}{bc\sqrt{b+c}}\)(2); \(\frac{1}{\sqrt{c^5+a^5}}\le\frac{1}{ca\sqrt{c+a}}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(VT=\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\)()

Xét \(\left(\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\right)^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\right)\)\(=\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\Rightarrow\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(2)

Từ (1) và (2) suy ra \(\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(đpcm)

Đẳng thức xảy ra khi \(a=b=c=\sqrt{3}\)

Gọi \(O\) là giao điểm của \(AC\)và \(BD\).

Theo đề ta có: \(\hept{\begin{cases}AC=8cm\\BD=6cm\end{cases}}\)

Theo tính chất của hình thoi ta có: \(\hept{\begin{cases}AO=OC=4cm\\BO=OD=3cm\end{cases}}\)

Áp dụng định lí Pitago trong \(\Delta AOB\) có:

\(AB^2=AO^2+OB^2\)

\(\Rightarrow AB=\sqrt{AO^2+OB^2}=\sqrt{4^2+6^2}\)

\(\Rightarrow AB=5cm\)

\(\Rightarrow S_{ABCD}=4AB=4.5=20cm\)

Vậy ...............

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]^2+4x\left(x+1\right)=12\)

\(\Leftrightarrow x^2\left(x+1\right)^2+4x\left(x+1\right)=12\)

\(\Leftrightarrow x^4+2x^3+x^2+4x^2+5x=12\)

\(\Leftrightarrow x^4+2x^3+5x^2+5x=12\)

\(\Leftrightarrow x^4+2x^3+5x^2+5x-12=0\)

\(\Leftrightarrow\left(x^3+3x^2+8x+12\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)=0\)

vi \(x^2+x+6\ne0\)nen:

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

sai dau sua ho toi

Đặt \(x^2+x=u\)

Phương trình trở thành \(u^2+4u=12\)

\(\Leftrightarrow u^2+4u-12=0\)

Ta có \(\Delta=4^2+4.12=64,\sqrt{\Delta}=8\)

\(\Rightarrow\orbr{\begin{cases}u=\frac{-4+8}{2}=2\\u=\frac{-4-8}{2}=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+x=2\\x^2+x=-6\end{cases}}\)

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

Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\)

+) \(x^2+x=-6\Leftrightarrow x^2+x+6=0\)

Mà \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

Vậy phương trình chỉ có 2 nghiệm là 1 và -2

\(ĐKXĐ:x\ne\pm1\)

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(x-1\right)}=0\)

\(\Leftrightarrow\frac{24+20\left(x^2-1\right)-\left(8x-1\right)\left(x-1\right)-\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2-11x+1=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)

ĐKXĐ: \(x\ne\pm1\)

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x-1\right)}+5=\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(1-x\right)}\)

\(\Leftrightarrow24\left(1-x\right)+20\left(x+1\right)\left(x-1\right)\left(1-x\right)=\left(8x-1\right)\left(x-1\right)\left(1-x\right)\)\(-\left(12x-1\right)\left(x+1\right)\left(1-x\right)\)

\(\Leftrightarrow4-4x+20x^2-20x^3=18x^2-20x^3+2x\)

\(\Leftrightarrow4-4x+20x^2=18x^2+2x\)

\(\Leftrightarrow4-4x+20x^2-18x^2-2x=0\)

1. He is used to------------ up early in the morning.
A. get B. getting C. have gotten D. got
2. The boy was kept indoors----------- naughty.
A. to be B. have been C. for being D. when he
3. I shall never forget-----------with you to Paris last year.
A. staying B. to staying C. to stay D. stayed
4. I am looking forward to -----------you.
A. having seen B. seeing C. to see D. all are wrong
5. I am always remember------------ off the lights before I leave my house.
A. turning B. to turn C. turned D. being turned