Bài 1:
\(\frac{4x+3}{3}-\frac{x-1}{6}>\frac{2x-3}{2}-\frac{x+2}{4}\)
=>\(\frac{4\left(4x+3\right)-2\left(x-1\right)}{12}>\frac{6\left(2x-3\right)-3\left(x+2\right)}{12}\)
=>4(4x+3)-2(x-1)>6(2x-3)-3(x+2)
=>16x+12-2x+2>12x-18-3x-6
=>14x+14>9x-24
=>5x>-38
=>x>-38/5
Bài 2:
a: ĐKXĐ: \(x^2+3x-6\ge0\)
=>\(x^2+3x+\frac94\ge6+\frac94=\frac{33}{4}\)
=>\(\left(x+\frac32\right)^2\ge\frac{33}{4}\)
=>\(\left[\begin{array}{l}x+\frac32\ge\frac{\sqrt{33}}{2}\\ x+\frac32\le-\frac{\sqrt{33}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{33}-3}{2}\\ x\le\frac{-\sqrt{33}-3}{2}\end{array}\right.\)
\(x^2+3x+4\cdot\sqrt{x^2+3x-6}=0\)
=>\(x^2+3x-6+4\cdot\sqrt{x^2+3x-6}+6=0\)
=>\(\left(\sqrt{x_{}^2+3x-6}+2\right)^2+2=0\) (vô lý)
=>x∈∅
b: \(\begin{cases}\frac{1}{x+y}+\frac{1}{x-y}=\frac58\\ \frac{1}{x-y}-\frac{1}{x+y}=\frac38\end{cases}\Rightarrow\begin{cases}\frac{1}{x+y}+\frac{1}{x-y}+\frac{1}{x-y}-\frac{1}{x+y}=\frac58+\frac38\\ \frac{1}{x+y}+\frac{1}{x-y}=\frac58\end{cases}\)
=>\(\begin{cases}\frac{2}{x-y}=\frac88=1\\ \frac{1}{x+y}=\frac58-\frac{1}{x-y}\end{cases}\Rightarrow\begin{cases}x-y=2\\ \frac{1}{x+y}=\frac58-\frac12=\frac18\end{cases}\Rightarrow\begin{cases}x-y=2\\ x+y=8\end{cases}\)
=>x=(2+8)/2=5; y=8-x=8-5=3
