a) Ta có: \(A=x^2+4x+10\)

\(=x^2+4x+4+6\)

\(=\left(x+2\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=x^2-2x+10\)

\(=x^2-2x+1+9\)

\(=\left(x-1\right)^2+9\ge9\forall x\)

Dấu '=' xảy ra khi x=2

c) Ta có: \(C=x^2-10x+10\)

\(=x^2-10x+25-15\)

\(=\left(x-5\right)^2-15\ge-15\forall x\)

Dấu '=' xảy ra khi x=5

d) Ta có: \(D=4x^2-4x+10\)

\(=4x^2-4x+1+9\)

\(=\left(2x-1\right)^2+9\ge9\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a) \(\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2=0+25=25\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)

b) \(8x^3-50x=0\)

\(2x\left(4x^2-25\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)

\(\Leftrightarrow\left(3x+7\right)\left(2x-5\right)=0\)

=>x=-7/3 hoặc x=5/2

\(2x\left(3x+7\right)-15x-35=0\\ \Rightarrow2x\left(3x+7\right)-\left(15x+35\right)=0\\ \Rightarrow2x\left(3x+7\right)-5\left(3x+7\right)=0\\ \Rightarrow\left(2x-5\right)\left(3x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)

\(\Leftrightarrow4x^2-12x-4x^2+9=-3\)

=>-12x=-12

hay x=1

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

\(4x^2-12x-4x^2+9+3=0\)

\(12-12x=0\\ \Rightarrow1-x=0\\ \Rightarrow x=1\)

a: \(\sqrt{252}+\dfrac{1}{3}\sqrt{63}-\sqrt{175}\)

\(=4\sqrt{7}+\sqrt{7}-5\sqrt{7}\)

=0

B = | 2x -10| + (5-x)² + 23

|2x-10| = 2.|x-5| \(\ge\) 0 ; (5-x)²\(\ge\) 0 \(\Rightarrow\) |2x-10| + (5-x)² + 23 \(\ge\) 23

B(min) = 23 dấu bằng xảy ra khi :

\(\left\{{}\begin{matrix}|x-5|=0\\5-x=0\end{matrix}\right.\)

\(\Rightarrow\) x = 5 

kết luận B(min) = 23 xảy ra khi x = 5