\(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}\ge\frac{1}{ab+1}\)

\(\Leftrightarrow ab^3-a^2b^2+a^3b-2ab+1\ge0\)

\(\Leftrightarrow ab\left(a-b\right)^2+\left(ab-1\right)^2\ge0\)đúng

Bạn bình phương P lên rồi tách hết ra

Sai đề rồi ạn ơi!

áp dụng bđt bunhiacopxki ta có:

\(\left(a+b+c\right)\left(1+1+1\right)>=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\Rightarrow3\cdot3=9>=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\)

\(\Rightarrow3>=\sqrt{a}+\sqrt{b}+\sqrt{c}\)

dấu = xảy ra khi a=b=c=1

vậy max A là 3 khi a=b=c=1

cách khác:

\(\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)

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

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

\(=x^6-1\)

minhf chịu

\(A=\left(2x+1\right)\left(x^2-x+2\right)-\left(2x-1\right)\left(x^2-2\right)-7x+5\)

    \(=2x\left(x^2-x+2\right)+x^2-x+2-2x\left(x^2-2\right)+\left(x^2-2\right)-7x+5\)

     \(=2x^3-2x^2+4x+x^2-x+2-2x^3+4x+x^2-2-7x+5\)

      \(=5\)

Vậy biểu thức k phụ thuộc vào x

P/s: Ủng hộ nha

\(\left(2x+1\right)\left(x^2-x+2\right)-\left(2x-1\right)\left(x^2-2\right)-7x+5\)
\(=2x^3-2x^2+4x+x^2-x+2-\left(2x^3-4x-x^2+2\right)\)\(-7x+5\)\(=2x^3-2x^2+4x+x^2-x+2-2x^3+4x+x^2-2-7x+5\)

\(=5\)không phụ thuộc vào x

a,\(=x^4-3x^3+3x^3-9x^2-4x^2+12x-12x+36\)

   \(=x^3\left(x-3\right)+3x^2\left(x-3\right)-4x\left(x-3\right)-12\left(x-3\right)\)

   \(=\left(x-3\right)\left(x^3+3x^2-4x-12\right)\)

    \(=\left(x-3\right)[x^2\left(x+3\right)-4\left(x+3\right)]\)

    \(=\left(x^2-9\right)\left(x^2-4\right)\)

a,    đktm:x khác 3    gọi a sao cho x-1=a

\(\Rightarrow A=\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{x-1-2}{\sqrt{a}-\sqrt{2}}=\frac{a-2}{\sqrt{a}-\sqrt{2}}=\frac{\left(\sqrt{a}-\sqrt{2}\right)\left(\sqrt{a}+\sqrt{2}\right)}{\sqrt{a}-\sqrt{2}}\)

\(=\sqrt{a}+\sqrt{2}=\sqrt{x-1}+\sqrt{2}\)

b,    \(A=\sqrt{x-1}+\sqrt{2}=\sqrt{4\left(2-3\right)-1}+\sqrt{2}=\sqrt{4\cdot-1-1}+\sqrt{2}=\sqrt{-5}+\sqrt{2}\)

vì \(\sqrt{-5}\)ko có nghĩa \(\Rightarrow\sqrt{-5}+\sqrt{2}\)ko có nghĩa \(\Rightarrow A\)ko có nghĩa khi \(x=4\left(2-3\right)\)

c       \(\sqrt{x-1}>=0\Rightarrow\sqrt{x-1}+\sqrt{2}>=\sqrt{2}\)

dấu = xảy ra khi \(x=1\)

\(\Rightarrow\)min A là \(\sqrt{2}\)khi x=1

a = __x - 3___

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