2.

C/m \(2a^2+2ab+1+a+b^2>0,\forall a,b\)

\(VT=a^2+2ab+b^2+a^2+a+1\)

\(=\left(a+b\right)^2+a^2+a+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(a+b\right)^2+\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrowđpcm\)

1. \(D=\left(x-1\right)^3-x\left(x-2\right)^2-x+2\)
\(D=\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left[x\left(x-2\right)+1\right]\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left[x^2-2x+1\right]\)
\(D=\left(x-1\right)^3-\left(x-2\right)\left(x-1\right)^2\)
\(D=\left(x-1\right)^2\left[\left(x-1\right)-\left(x-2\right)\right]\)
\(D=\left(x-1\right)^2\left[\left(x-1-x+2\right)\right]\)
\(D=\left(x-1\right)^2\left[1\right]\)
\(D=\left(x-1\right)^2\)
\(D=0\Rightarrow\left(x-1\right)^2=0\)
\(x=1\)

KL: x = 1

Vì \(x\in N\) nên x>=0
\(\Rightarrow x+2>=2>0\)
\(x+4>=4>0\)
\(x+3>=3>0\)
Do đó \(\left(x+2\right)\left(x+4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)

KL: x = 3

\(f\left(x\right)=\dfrac{\sqrt{x^2-2}}{\sqrt{x^2-3}-1}+\dfrac{1}{\sqrt{x^2+1}+1}\)

(1) )\(x^2-2\ge0\Rightarrow\left|x\right|\ge\sqrt{2}\)

(2) \(x^2-3\ge0\Rightarrow\left|x\right|\ge\sqrt{3}\)

(3) \(\sqrt{x^2-3}-1\ne0\Rightarrow\left|x^2-3\right|\ne1\Rightarrow\left|x\right|\ne2\)

(4) \(x^2+1\ge0\Rightarrow\forall x\)

(5) \(\sqrt{x^2+1}+1\ne0\Rightarrow\forall x\)

Từ (1),(2),(3),(4) và (5):

\(\left|x\right|\ge\sqrt{3}\) và \(x\ne\left|2\right|\)

KL: \(x\le-\sqrt{3}\) và \(x\ne-2\)

Hoặc \(x\ge\sqrt{3}\) và \(x\ne2\)

Cái này chỉ cần tiềm trên mạng là ok

\(A=1367+5472\)
\(=1000+300+67+5000+400+72\)

\(=1000+300+67+5000+400+62+10\)
\(=10+62+67+300+400+1000+5000\)
\(B=5377+1462\)
\(=5000+300+77+1000+400+62\)
\(=5000+300+67+10+1000+400+62\)
\(=10+62+67+300+400+1000+5000\)

KL: A = B

Đề sai 100%

Tại sao DE =CF

Một hình thang cân bất kì

Không thoả mãn đề bài rồi => đề sai