\(C=4x-x^2+1\\ =-\left(x^2-4x-1\right)\\ =-\left[\left(x^2-4x+4\right)-5\right]\\ =-\left(x-2\right)^2+5\le5\forall x\in R\)

\(Dấu''=''\ xảy\ ra\ khi :(x-2)^2=0<=>x=2\)

\(Vậy\ GTLN\ của\ biểu\ thức\ C\ là:5\ khi\ x=2\)

\(D=3-10x^2-4xy-4y^2\\ =-\left(x^2+4xy+4y^2+9x^2-3\right)\\ =-\left[\left(x+2y\right)^2+9x^2-3\right]\\ =-\left[\left(x+2y\right)^2+9x^2\right]+3\le3\forall x;y\in R\)

\(Dấu\ ''=''\ xảy\ ra\ khi:\) \(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\9x^2=0\end{matrix}\right.< =>x=0;y=0\)

\(Vậy\ GTLN\ của\ biểu\ thức\ D\ là : 3\ khi\ x=y=0\)

\(M=25x^2-20x+7\\ =\left(5x\right)^2-2.5x.2+2^2+3\\ =\left(5x-2\right)^2+3>=3>0\left(DPCM\right)\)

\(N=9x^2-6xy+2y^2+1\\ =\left(9x^2-6xy+y^2\right)+y^2+1\\ =\left(3x-y\right)^2+y^2+1>=1>0\left(DPCM\right)\)

\(A=x^2+12x+39\\ =\left(x^2+12x+36\right)+3\\ =\left(x+6\right)^2+3>=3\forall x\in R\)

\(Dấu ''=''\ xảy\ ra\ khi : (x+6)^2=0<=>x=-6\)

\(Vậy\ GTNN\ của\ biểu\ thức\ A\ là :3\ khi\ x=-6\)

\(B=9x^2-12x\\ =\left(3x\right)^2-2.3x.2+2^2-2^2\\ =\left(3x-2\right)^2-4>=-4\)

\(Dấu ''=''\ xảy\ ra\ khi:(3x-2)^2=0<=>x=2/3\)

\(Vậy\ GTNN\ của\ biểu\ thức\ B\ là : -4\ khi\ x=2/3\)

\(A=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\\ =\left[\left(x^2+1\right)+x\right].\left[\left(x^2+1\right)-x\right].\left(x^4-x^2+1\right)\\ =\left[\left(x^2+1\right)^2-x^2\right].\left(x^4-x^2+1\right)\\ =\left(x^4+2x^2+1-x^2\right).\left(x^4-x^2+1\right)\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =x^8+2x^4+1-x^4\\ =x^8+x^4+1\)

\(Ta\ thấy:\) \(x^8;x^4>=0\forall x\in R\\ =>x^8+x^4+1>=1\)

\(Hay : A>0(DPCM)\) 

\(x+y=-7< =>\left(x+y\right)^2=\left(-7\right)^2\\ < =>x^2+2xy+y^2=49\\ < =>x^2-2xy+y^2=49-4xy\\ < =>\left(x-y\right)^2=49-4.12\\ < =>\left(x-y\right)^2=1\)

\(=>x-y=\pm1\)

\(Vì : x>y=>x-y>0\)

\(Do\ đó, x-y=1\)

\(A=x^2-y^2=\left(x-y\right)\left(x+y\right)\)

\(Thay \ x-y=1;x+y=-7\ vào\ biểu\ thức\ trên,\ ta \ được :\)

\(A=1.\left(-7\right)=-7\)

\(Vậy\ giá\ trị\ biểu\ thức:\ x^2-y^2\ =-7\ khi :x+y=-7;xy=12(x>y)\)

\(3\sqrt{x+3}+\sqrt{x^2-9}=0\left(ĐK:x+3>=0;x^2-9>=0\right)\\ < =>3\sqrt{x+3}+\sqrt{\left(x-3\right)\left(x+3\right)}=0\\ < =>\sqrt{x+3}.\left(3+\sqrt{x-3}\right)=0\)

\(=>\left[{}\begin{matrix}\sqrt{x+3}=0\\3+\sqrt{x-3}=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-3\\\sqrt{x-3}=-3< 0\left(KTM\right)\end{matrix}\right.< =>x=-3\left(TMDK\right)=>S=\left\{-3\right\}\)

\(Gọi\ số\ bé\ là : \) \(\overline{x}\)

\(=> Số\ lớn\ là : \)\(\overline{x2}\)

\(Ta\ có:\) \(\overline{x}+\overline{x2}=8197\\ \overline{x}+\overline{x0}+2=8197\\ \overline{x}+\overline{x}\times10=8197-2\\ \overline{x}\times11=8195\\ \overline{x}=\dfrac{8195}{11}=745\)

\(Vậy\ 2\ số\ phải\ tìm\ là : 745;7452\)

\(\overline{abc5}-\overline{abc}=2129\\ \overline{abc0}+5-\overline{abc}=2129\\ \overline{abc}\times10-\overline{abc}=2129-5\\ \overline{abc}\times9=2124\\ \overline{abc}=\dfrac{2124}{9}=236\)

\(Gọi\ số\ thứ\ nhất\ là : x\)

\(Vì\ hiệu\ hai\ số\ là: 22=>Số\ thứ\ hai\ là :x+22\)

\(Ta\ có : x+x+22+22=116\)

\(2\times x+44=116\\ 2\times x=116-44=72\\ x=\dfrac{72}{2}=36\\ =>x+22=36+22=58\)

\(Vậy\ hai\ số\ đó\ là :36;58\)

\(y^2-24y+144=y^2-2.y.12+12^2=\left(y-12\right)^2\\ y^2-\left(7x\right)^2=\left(y-7x\right)\left(y+7x\right)\\ y^2+14y+49=y^2+2.y.7+7^2=\left(y+7\right)^2\\ 25x^2-64y^2=\left(5x\right)^2-\left(8y\right)^2=\left(5x-8y\right)\left(5x+8y\right)\)