a) 575-(6x+70)=455

6x+70=575-455

6x+70=120

6x=120-70

6x=50

x=50:6

x=\(\dfrac{25}{3}\) \(\notinℕ\)

Vậy không có giá trị tự nhiên x thỏa mãn đề

b) 315+(125-x)=435

125-x=435-315

125-x=120

x=125-120

x=5 (nhận)

c) 3x+28=88

3x=88-28

3x=60

x=60:3

x=20

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

Bạn xem lại đề nhé.

\(x^2+2x+10=\left(x^2+2x+1\right)+9\\ =\left(x+1\right)^2+9\ge9>0\forall x\inℝ\left(Vì:\left(x+1\right)^2\ge0\forall x\inℝ\right)\)

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+1+\dfrac{x+6}{2004}+1+\dfrac{x+7}{2003}+1=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

b) \(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}+\dfrac{3}{97.100}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

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

\(\forall x\inℝ:x^2\ge0\Rightarrow5x^2\ge0\\ \Rightarrow B=5x^2+5\ge5\\ \Rightarrow Min_B=5\)

Dấu = xảy ra khi: \(x^2=0\Leftrightarrow x=0\)

\(\dfrac{4}{10000}\)

\(B=4x^2-4x+1+x^2+4x+4\\ =5x^2+5\ge5\forall x\inℝ\) (Vì: \(\forall x\inℝ\Rightarrow x^2\ge0\))

Dấu = xảy ra khi: \(x^2=0\Leftrightarrow x=0\)

\(Min_B=5tạix=0\)

\(A=x^2-3x+5\\ =x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\dfrac{9}{4}+5\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\inℝ\)

Dấu = xảy ra khi: \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

\(Min_A=\dfrac{11}{4}tạix=\dfrac{3}{2}\)

\(B=-x^2-2y^2+2xy-2x+10y-40\\ =-\left(x^2-2xy+y^2\right)-y^2-2x+10y-40\\ =-\left(x-y\right)^2-2\left(x-y\right)-1-y^2+8y-39\\ =-\left[\left(x-y\right)^2+2.\left(x-y\right).1+1^2\right]-\left(y^2-8y+16\right)-23\\ =-\left(x-y+1\right)^2-\left(y-4\right)^2-23\le-23< 0\left(DPCM\right)\)

Dấu = xảy ra khi: \(x-y+1=y-4=0< \\ \Leftrightarrow y=4,x=3\)