\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(\left(a\right):\sqrt{16x}=8\left(ĐK:x\ge0\right)\\ < =>4\sqrt{x}=8\\ < =>\sqrt{x}=2\\ < =>x=4\left(TMDK\right)\)

\(\left(b\right):\sqrt{4x}=\sqrt{5}\left(ĐK:x\ge0\right)\\ < =>4x=5\\ < =>x=\dfrac{5}{4}\left(TMDK\right)\)

\(\left(c\right):\sqrt{9\left(x-1\right)}=21\left(ĐK:x\ge1\right)\\ < =>3\sqrt{x-1}=21\\ < =>\sqrt{x-1}=7\\ < =>x-1=49\\ < =>x=50\left(TMDK\right)\)

\(\left(d\right):\sqrt{4\left(1-x\right)^2}-6=0\left(ĐK:x\in R\right)\\ < =>2\left|1-x\right|=6\\ < =>\left|1-x\right|=3\\ =>\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\left(TMDK\right)\)

\(\left(a\right):A=-x^2+4x+2=-\left(x^2-4x-2\right)\\ =-\left[\left(x^2-4x+4\right)-6\right]\\ =-\left(x-2\right)^2+6\le6\forall x\)

\(Dấu''=''\ xảy\ ra\ khi:x-2=0<=>x=2\)

\(Vậy\ GTLN\ của\ A\ là :6<=>x=2\)

\(\left(b\right):B=x-x^2+6=-\left(x^2-x-6\right)\\ =-\left[\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{25}{4}\right]\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

\(Dấu''=''\ xảy\ ra\ khi :x-1/2=0<=>x=1/2\)

\(Vậy\ GTLN\ của\ B\ là :25/4<=>x=1/2\)

\(\left(c\right):C=-4y^2+8y-12=-\left(4y^2-8y+12\right)\\= -\left[\left(4y^2-8y+4\right)+8\right]\\ =-\left(2y-2\right)^2-8\le-8\forall y\)

\(Dấu''=''\ xảy\ ra\ khi:2y-2=0<=>y=1\)

\(Vậy\ GTLN\ của\ C\ là:-8<=>y=1\)

\(Q=x^2+y^2-4x-y+7\\ =\left(x^2-4x+4\right)+\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{11}{4}\\ =\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x,y\)

\(Dấu''=''\ xảy\ ra\ khi:\)

\(x-2=y-\dfrac{1}{2}=0< =>x=2;y=\dfrac{1}{2}\)

\(Vậy\ GTNN\ của\ Q\ là :11/4<=>x=2;y=1/2\)

\(\dfrac{x}{3}=\dfrac{y}{4}=>\dfrac{1}{5}.\dfrac{x}{3}=\dfrac{1}{5}.\dfrac{y}{4}=>\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\\ \dfrac{y}{5}=\dfrac{z}{7}=>\dfrac{1}{4}.\dfrac{y}{5}=\dfrac{1}{4}.\dfrac{z}{7}=>\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\\ \left(1\right);\left(2\right)=>\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{124}{62}=2\\ =>x=2.15=30;y=2.20=40;z=2.28=56\)

\(\left(a\right):9-4x^2=3^2-\left(2x\right)^2=\left(3-2x\right)\left(3+2x\right)\)

\(\left(b\right):16x^2-25=\left(4x\right)^2-5^2=\left(4x-5\right)\left(4x+5\right)\)

\(\left(c\right):a^4-16=\left(a^2\right)^2-4^2=\left(a^2-4\right)\left(a^2+4\right)\\ =\left(a^2-2^2\right)\left(a^2+4\right)=\left(a-2\right)\left(a+2\right)\left(a^2+4\right)\)

\(\left(d\right):\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b+1\right)\)

\(\left(4x+2\right)\left(x+3\right)-\left(2x-3\right)^2\\ =4x^2+2x+12x+6-\left(4x^2-12x+9\right)\\ =4x^2+14x+6-4x^2+12x-9\\ =26x-3\)

\(2^{150}=2^{3.50}=\left(2^3\right)^{50}=8^{50}\\ 3^{100}=3^{2.50}=\left(3^2\right)^{50}=9^{50}\)

\(Vì : 8^{50}<9^{50}=>2^{150}<3^{100}\)

\(3\sqrt{4x+4}-2\sqrt{9x+9}+\sqrt{x+1}=\sqrt{2}+1\left(x>=-1\right)\\ < =>3.\sqrt{4}.\sqrt{x+1}-2.\sqrt{9}.\sqrt{x+1}+\sqrt{x+1}=\sqrt{2}+1\\ < =>\left(3.2-2.3+1\right).\sqrt{x+1}=\sqrt{2}+1\\ < =>\sqrt{x+1}=\sqrt{2}+1\\ < =>x+1=\left(\sqrt{2}+1\right)^2=3+2\sqrt{2}\\ < =>x=2+2\sqrt{2}\left(TMDK\right)\)

\(\left(b\right):\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\left(ĐK:x\ne\left\{-\dfrac{7}{5};-\dfrac{1}{5}\right\}\right)\\ =>\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\\ =>15x^2+10x+3x+2=15x^2-5x+21x-7\\ =>15x^2+13x+2=15x^2+16x-7\)

\(=>15x^2-15x^2+16x-13x=7+2\\ =>3x=9\\ =>x=3\)

\(\left(e\right):x^2+x=0\\ =>x\left(x+1\right)=0\\ =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)