Bài 2:

\(a.\dfrac{5x-3}{x+2}-4=\dfrac{6}{x+2}\left(x\ne-2\right)\\ \Leftrightarrow\dfrac{5x-3}{x+2}-\dfrac{6}{x+2}=4\\ \Leftrightarrow\dfrac{5x-3-6}{x+2}=4\\ \Leftrightarrow\dfrac{5x-9}{x+2}=4\\ \Leftrightarrow5x-9=4\left(x+2\right)\\ \Leftrightarrow5x-9=4x+8\\ \Leftrightarrow5x-4x=8+9\\ \Leftrightarrow x=17\left(tm\right)\)

\(b.\dfrac{1}{x}-\dfrac{x+2}{x-2}=\dfrac{-2}{x\left(x-2\right)}\left(x\notin\left\{0;2\right\}\right)\\ \Leftrightarrow\dfrac{x-2}{x\left(x-2\right)}-\dfrac{x\left(x+2\right)}{x\left(x-2\right)}=\dfrac{-2}{x\left(x-2\right)}\\ \Leftrightarrow x-2-x\left(x+2\right)=-2\\ \Leftrightarrow x-2-x^2-2x=-2\\ \Leftrightarrow-x^2-x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

\(c.\dfrac{7}{x-5}-2=\dfrac{3}{5-x}\left(x\ne5\right)\\ \Leftrightarrow\dfrac{7}{x-5}-2-\dfrac{3}{x-5}=0\\ \Leftrightarrow\dfrac{7}{x-5}+\dfrac{3}{x-5}=2\\ \Leftrightarrow\dfrac{10}{x-5}=2\\ \Leftrightarrow x-5=\dfrac{10}{2}=5\\\Leftrightarrow x=5+5\\ \Leftrightarrow x=10\left(tm\right)\)

Bài 3:

a: ĐKXĐ: \(x\notin\left\{-2;2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x-1}{x-2}=\dfrac{-3x+2}{x^2-4}\)

=>\(\dfrac{x\left(x-2\right)-\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3x+2}{\left(x-2\right)\left(x+2\right)}\)

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

=>-3x+2=-3x+2

=>0x=0(luôn đúng)

Vậy: \(x\in R\backslash\left\{-2;2\right\}\)

b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x-1}{x+2}+\dfrac{x+1}{x-2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)+\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

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

=>0x=0(luôn đúng)

vậy: \(x\in R\backslash\left\{2;-2\right\}\)

\(b)\left\{{}\begin{matrix}x-\dfrac{y}{2}=\dfrac{1}{2}\\\dfrac{x}{3}-2y=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-6y=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\2x-12y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=11\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x=1+1=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

\(c)\left\{{}\begin{matrix}5x-0,7y=1\\-10x+1,4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\5x-0,7y=1\end{matrix}\right.\) 

=> Hpt vô số nghiệm 

j

j

h

 Xét \(f\left(x\right)=VT=x^2+y^2+xy-3x-3y+3\)

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

 Có \(\Delta=\left(y-3\right)^2-4\left(y^2-3y+3\right)\)

\(=y^2-6y+9-4y^2+12y-12\)

\(=-3y^2+6y-3\)

\(=-3\left(y-1\right)^2\le0\) với mọi \(y\inℝ\)

 Mà \(f\left(x\right)\) có hệ số cao nhất bằng \(1>0\) nên từ đây có \(VT=f\left(x\right)\ge0\)

 Dấu "=" xảy ra khi \(y=1\). Khi đó \(\Delta=0\) nên pt \(f\left(x\right)=0\) có nghiệm kép \(\Leftrightarrow\) \(x=\dfrac{-\left(y-3\right)}{2}=1\).

 Ta có đpcm.

h

sh

a, Ta có: \(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(SO_3+H_2O\rightarrow H_2SO_4\)

_____0,1____________0,1 (mol)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

______0,1_______0,2 (mol)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(\dfrac{4}{\sqrt{5}+\sqrt{3}}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}-\sqrt{2^2\cdot5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-2\sqrt{5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{2}-2\sqrt{5}\\ =2\left(\sqrt{5}-\sqrt{3}\right)-2\sqrt{5}\\ =2\sqrt{5}-2\sqrt{3}-2\sqrt{5}\\ =-2\sqrt{3}\)