viết đoạn văn nhé

a) Ta có: \(A=4x^2+4x+11\)

        \(\Rightarrow A=4x^2+2x+2x+11\)

        \(\Rightarrow A=2x.\left(2x+1\right)+\left(2x+1\right)+10\)

        \(\Rightarrow A=\left(2x+1\right).\left(2x+1\right)+10\)

        \(\Rightarrow A=\left(2x+1\right)^2+10\)

  Ta lại có: \(\left(2x+1\right)^2\ge0\forall x\inℝ\)

             \(\Rightarrow A\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\)

                        \(\Rightarrow2x+1=0\)

                        \(\Rightarrow2x=-1\)

                        \(\Rightarrow x=\frac{-1}{2}\)

      Vậy \(A_{min}=10\Leftrightarrow x=\frac{-1}{2}\)

a, a^3 + b^3=(a + b)^3 - 3a²b - 3ab²=(a + b)^3 - 3ab(a + b)

b, a^3 + b^3 + c^3 - 3abc= (a + b)^3 + c³ - 3ab(a + b)-3abc

=(a + b + c)\([\)(a + b)²- (a + b)c +c²\(]\)- 3ab(a + b + c)

=(a + b + c)(a² + 2ab + b² - ac - bc + c² - 3ab)

=(a + b + c)(a² + b² + c² - ab - bc- ca)

\(a.A=100^2-99^2+98^2-97^2+...+2^2-1\)

        \(=100+99+98+97+...+2+1\)

         \(=\frac{\left(100+1\right).100}{2}=5050\)(công thức tính dãy số hạng)

\(b.B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

         \(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

           \(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

           \(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{64}+1\right)+1\)

            \(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

             \(=2^{4096}-1+1\)

              \(=2^{4096}\)

\(c.\)Đặt\(a+b=d\)

       Thay vào \(C\)ta được:

\(C=\left(d+c\right)^2+\left(d-c\right)^2-2d^2\)

     \(=d^2+2dc+c^2+d^2-2dc+c^2-2d^2\)

      \(=2c^2\)

Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(\frac{3}{2}\right)^2\Leftrightarrow a^2+b^2+c^2\ge\frac{3}{4}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=b=c\\a+b+c=\frac{3}{2}\end{cases}\Leftrightarrow}a=b=c=\frac{1}{2}\)

\(\Leftrightarrow\)(x²+1)²+x(x²+1)+2x(x²+1)+2x²=0

\(\Leftrightarrow\)(x²+1)(x²+1+x)+2x(x²+1+x)=0

\(\Leftrightarrow\)(x²+1+x)(x²+1+2x)=0

\(\Leftrightarrow\)(x²+x+1)(x+1)²=0

Vì x²+x+1=(x²+x+\(\frac{1}4\))+\(\frac{3}4\)=(x+\(\frac{1}2\))²+\(​​​​\frac{3}{4}\)\(\ge\)\(​​​​\frac{3}{4} \)>0 nên x+1=0, x=1.

\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

\(x^4+2x^2+1+3x^3+3x+2x^2=0\)

\(x^4+3x^3+4x^2+3x+1=0\)

\(x^4+2x^3+x^3+2x^2+2x^2+x+2x+1=0\)

\(\left(x+1\right)\left(x^3+2x^2+1+2x\right)=0\)

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

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

\(\orbr{\begin{cases}x+1=0\Rightarrow x=\left(-1\right)\\x^2+x+1=0\end{cases}}\)

x^2 +x +1 =0 (vô lí)

\(\Rightarrow x=\left(-1\right)\)

\(\frac{x^2+2}{x^2+1}=\frac{x^2+1+1}{x^2+1}=\frac{x^2+1}{x^2+1}+\)\(\frac{1}{x^2+1}\)\(=1+\frac{1}{x^2+1}\)

\(M_{min}=1\Leftrightarrow\frac{1}{x^2+1}\)nhỏ nhất \(\Leftrightarrow x^2+1\)lớn  nhất

kết luận: sai đề, nếu tìm giá trị lớn nhất thì \(x^2+1\)nhỏ nhất nên x = 0 chứ làm sao tìm được giá trị lớn nhất của \(x^2+1\)được???

xin lỗi bạn nha phải là x^2-2