Đặt A = ( 3 + 1 )( 3² + 1 )...( 3⁶⁴ + 1 ) + 1

=> 2A = 2( 3 + 1 )( 3² + 1 )...( 3⁶⁴ + 1 ) + 2

= ( 3 - 1 )( 3 + 1 )( 3² + 1 )...( 3⁶⁴ + 1 ) + 2

= ( 3² - 1 )( 3² + 1 )...( 3⁶⁴ + 1 ) + 2

= ( 3⁶⁴ - 1 )( 3⁶⁴ + 1 ) + 2

= 3¹²⁸ - 1 + 2 = 3¹²⁸ + 1

=> A = ( 3¹²⁸ + 1 )/2

ngu moi di hoi

\(2x+5x=12-9x\)

\(\Rightarrow7x=12-9x\)

\(\Rightarrow7x+9x=12\)

\(\Rightarrow16x=12\)

\(\Rightarrow x=\frac{3}{4}\)

Ta có :\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

=> \(a\left(\frac{a}{b+c}\right)+b\left(\frac{b}{a+c}\right)+c\left(\frac{c}{a+b}\right)=0\)

=> \(a\left(\frac{a}{b+c}+1-1\right)+b\left(\frac{b}{a+c}+1-1\right)+c\left(\frac{c}{a+b}+1-1\right)=0\)

=> \(a\left(\frac{a+b+c}{b+c}-1\right)+b\left(\frac{a+b+c}{a+c}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)=0\)

=> \(a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{a+c}-b+c.\frac{a+b+c}{a+b}-c=0\)

=> \(\left(a+b+c\right).\frac{a}{b+c}+\left(a+b+c\right).\frac{b}{a+c}+\left(a+b+c\right).\frac{c}{a+b}-\left(a+b+c\right)=0\)

=> \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}-1\right)=0\)

=> \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}-1=0\left(\text{Vì }a+b+c\ne0\right)\)

=> \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)(đpcm)

\(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra <=> x = 3

Vậy MinA = 1

\(B=5x^2-10x+3=5\left(x^2-2x+1\right)-2=5\left(x-1\right)^2-2\ge-2\forall x\)

Dấu "=" xảy ra <=> x = 1

Vậy MinB = -2

\(C=2x^2+8x+y^2-10y+43=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)

Dấu "=" xảy ra <=> x = -2 ; y = 5

Vậy MinC = 10

\(A=x^2-6x+10\)

\(=\left(x^2-6x+9\right)+1\)

\(=\left(x-3\right)^2+1\ge1\forall x\)

Dấu"=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)

Vậy \(Min_A=1\Leftrightarrow x=3\)

b,\(B=5x^2-10x+3\)

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

\(=5\left(x-1\right)^2-2\ge-2\forall x\)

Dấu"=" xảy ra khi \(x-1=0\Leftrightarrow x=1\)

Vậy \(Min_B=-2\Leftrightarrow x=1\)

c,\(C=2x^3+8x+y^2-10+43\)

\(=2x^2+8x+8+y^2-10y+25+10\)

\(=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10\)

\(=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)

Dấu"=" xảy ra khi \(\orbr{\begin{cases}x+2=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=5\end{cases}}}\)

Vậy \(Min_C=10\Leftrightarrow x=-2;y=5\)

x⁸ + 3x⁴ + 4

= ( x⁸ + 4x⁴ + 4 ) - x⁴

= ( x⁴ + 2 )² - ( x² )²

= ( x⁴ - x² + 2 )( x⁴ + x² + 2 )

\(=x^8+4x^4+4-x^4\)

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

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

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

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

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

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

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)

\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)

\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)

\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow2x+4=2x+3\)

\(\Leftrightarrow0x=-1\)(vô nghiệm)

Vậy phương trình vô nghiệm.

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow2x+7=-10\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)