\(2\left(2x-5\right)^2-\left(2x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=70\)

\(2\left(4x^2-20x-25\right)-\left(4x^2-12x+9\right)-\left(4x^2-1\right)=70\)

\(8x^2-40x-50-4x^2+12x-9-4x^2+1=70\)

\(-28x-128=0\)

\(28x=-128\)

\(x=\frac{32}{7}\)

Đặt \(2x^2-1=a\)

\(\Rightarrow\frac{a}{x}+\frac{5x}{a-x}=-7\)

\(\Leftrightarrow2x^2-6ax-a^2=0\)

Đặt \(a=tx\)

\(\Rightarrow2x^2-6tx^2-t^2x^2=0\)

\(\Leftrightarrow2-6t-t^2=0\)

Làm nốt nha

\(m>n\Rightarrow m=n+p\left(p>0\right)\)

\(\Rightarrow x^m=x^n\cdot x^p\)mà \(x< 1\Rightarrow x^m=x^n\cdot x^p< x^n\cdot1^p=x^n\cdot1=x^n\Rightarrow x^m< x^n\)(đpcm)

C=(x+3)²-3

Ta thấy: \(\left(x+3\right)^2\ge0\)

\(\Rightarrow\left(x+3\right)^2-3\ge-3\)

Min_C= -3 khi x = -3

(x+3)²>=0 với mọi x

(x+3)²-3>=-3

C>=-3

già trị nhỏ nhất của C=-3 <=> x+3=0<=>x=-3

Ta có :

\(2x-x^2\)

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

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

Vì \(\left(x-1\right)^2\ge\forall x\)

\(\Leftrightarrow-\left(x-1\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x-1\right)^2+1\le1\)

Vậy ĐPCM

\(\left(x+5y\right)^2-\left(2x-7y\right)^2\)

\(=\left(x+5y-2x+7y\right)\left(x+5y+2x-7y\right)\)

\(=\left(-x+12y\right)\left(3x-2y\right)\)

\(\left(x+5y\right)^2-\left(2x-7y\right)^2\)

\(=x^2+10xy+25y^2-4x^2+28xy+49y^2\)

\(=\left(x^2-4x^2\right)+\left(10xy+28xy\right)+\left(25y^2+49y^2\right)\)

\(=-3x^2+38xy+74y^2\)