\(a,PT\left(x\ne1;-1\right).\Leftrightarrow\dfrac{x^2+2x+1+x^2-2x+1-4}{\left(x-1\right)\left(x+1\right)}=0.\\ \Rightarrow x^2+2x+1+x^2-2x+1-4=0.\\ \Leftrightarrow2x^2-2=0.\\ \Leftrightarrow x^2-1=0.\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0.\)

\(\Leftrightarrow x=\pm1\left(koTM\right).\)

\(b,x^2+1\ge2x.\\ \Leftrightarrow x^2-2x+1\ge0.\\ \Leftrightarrow\left(x-1\right)^2\ge0.\\ \Leftrightarrow\forall x\in R.\)

\(c,x\left(x-1\right)\ge x-1.\\ \Leftrightarrow x\left(x-1\right)-\left(x-1\right)\ge0.\\ \Leftrightarrow\left(x-1\right)\left(x-1\right)\ge0.\\ \Leftrightarrow\left(x-1\right)^2\ge0.\\ \Leftrightarrow\forall x\in R.\)

Ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o.\)

Mà \(\widehat{B}=\widehat{A}+10^o.\\ \widehat{C}=\widehat{B}+10^o=\widehat{A}+10^o+10^o=\widehat{A}+20^o.\\ \widehat{D}=\widehat{C}+10^o=\widehat{A}+20^o+10^o=\widehat{A}+30^o.\)

\(\Rightarrow\widehat{A}+\widehat{A}+10^o+\widehat{A}+20^o+\widehat{A}+30^o=360^o.\\ \Leftrightarrow4\widehat{A}=300^o.\\ \Leftrightarrow\widehat{A}=75^o.\\ \Rightarrow\widehat{B}=85^o.\\ \widehat{C}=95^o.\\ \widehat{D}=105^o.\)

\(A=x^2-10x.\\ A=x^2-2.5x+25-25.\\ A=\left(x-5\right)^2-25.\)

Ta có: \(\left(x-5\right)^2\ge0.\Leftrightarrow\left(x-5\right)^2-25\ge-25.\\ \Rightarrow A\ge-25.\)

Dấu = xảy ra \(\Leftrightarrow A=-25.\Leftrightarrow x=5.\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne2;x\ne4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2+\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=0.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\) (TM).

Bài 4:

\(1,A=\left\{1;2;3;4;a;b;c\right\}.\\ B=\left\{3;4;b;c\right\}.\\ 2,M=\left\{1;2;a\right\}.\\ 3,N=\left\{4;3;b;c\right\}.\)

Bài 5:

\(1.A=\left\{a;b\right\};B=\left\{c;i\right\}.\\2. C=\left\{c\right\}.\)

\(c,5\sqrt{12x}-4\sqrt{3x}+2\sqrt{48x}=14.\\ \Leftrightarrow10\sqrt{3x}-4\sqrt{3x}+8\sqrt{3x}=14.\\ \Leftrightarrow14\sqrt{3x}=14.\\ \Leftrightarrow\sqrt{3x}=1.\\ \Leftrightarrow3x=1.\\ \Leftrightarrow x=\dfrac{1}{3}.\)

\(d,2\sqrt{x-1}+\dfrac{3}{2}\sqrt{4x-4}-\dfrac{1}{3}\sqrt{9x-9}=12.\\ \Leftrightarrow2\sqrt{x-1}+\dfrac{3}{2}\sqrt{4\left(x-1\right)}-\dfrac{1}{3}\sqrt{9\left(x-1\right)}=12.\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-\sqrt{x-1}=12.\\ \Leftrightarrow4\sqrt{x-1}=12.\\ \Leftrightarrow\sqrt{x-1}=3.\\ \Leftrightarrow x-1=9.\\ \Leftrightarrow x=10.\)

\(e,\dfrac{2}{5}\sqrt{25x+25}-3\sqrt{x+1}+\dfrac{5}{2}\sqrt{4x+4}=8.\\ \Leftrightarrow\dfrac{2}{5}\sqrt{25\left(x+1\right)}-3\sqrt{x+1}+\dfrac{5}{2}\sqrt{4\left(x+1\right)}=8.\\ \Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}+5\sqrt{x+1}=8.\\ \Leftrightarrow4\sqrt{x+1}=8.\\ \Leftrightarrow\sqrt{x+1}=2.\\ \Leftrightarrow x+1=4.\\ \Leftrightarrow x=3.\)

\(g,3\sqrt{2x}+4\sqrt{8x}-\sqrt{18x}=10.\\ \Leftrightarrow g,3\sqrt{2x}+8\sqrt{2x}-3\sqrt{2x}=10.\\ \Leftrightarrow8\sqrt{2x}=10.\\ \Leftrightarrow\sqrt{2x}=\dfrac{5}{4}.\\ \Leftrightarrow2x=\dfrac{25}{16}.\\ \Leftrightarrow x=\dfrac{25}{32}.\)

Ta có: \(MB=\dfrac{1}{4}AB=\dfrac{1}{4}.8=2\left(cm\right).\)

\(AM=AB-BM.\\ =8-2=6\left(cm\right).\)

Xét \(\Delta AMD\) vuông tại A:

\(MD^2=AD^2+AM^2\left(Pytago\right).\\ \Rightarrow MD=\sqrt{AD^2+AM^2}.\\ \Rightarrow MD=\sqrt{8^2+6^2}=10\left(cm\right).\)

\(\Delta DNC:MB//DC\) \((\)cùng \(\perp BC).\)

\(\Rightarrow\dfrac{MB}{DC}=\dfrac{MN}{ND}\left(Talet\right).\\ \Leftrightarrow\dfrac{MB}{DC}=\dfrac{ND-MD}{ND}.\\ \Rightarrow\dfrac{2}{8}=\dfrac{ND-10}{ND}.\\ \Rightarrow\dfrac{1}{4}=\dfrac{ND-10}{ND}.\)

\(\Rightarrow4ND-40-ND=0.\\ \Leftrightarrow3ND=40.\\ \Leftrightarrow ND=\dfrac{40}{3}\left(cm\right).\)