a. \(5x-7\ge3\left(x-2\right)+x\)
\(\Leftrightarrow5x-7\ge3x-6+x\)
\(\Leftrightarrow x\ge1\)
Vậy: Bất phương trình có tập nghiệm \(S=\left\{x|x\ge1\right\}\)
b. \(3\left(4x+1\right)-2\left(5x+2\right)>8x-2\)
\(\Leftrightarrow12x+3-10x-4>8x-2\)
\(\Leftrightarrow-6x>-1\)
\(\Leftrightarrow x< \dfrac{1}{6}\)
Vậy: ...
c. \(1+x-\dfrac{x-3}{4}\ge\dfrac{x+1}{4}-\dfrac{x-2}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+7\right)}{12}\ge\dfrac{3\left(x+1\right)-4\left(x-2\right)}{12}\)
\(\Leftrightarrow3\left(3x+7\right)\ge3\left(x+1\right)-4\left(x-2\right)\)
\(\Leftrightarrow9x+21\ge3x+3-4x+8\)
\(\Leftrightarrow10x\ge-10\)
\(\Leftrightarrow x\ge-1\)
Vậy: ...
a,
\(5x-7\ge3\left(x-2\right)+x\)
\(\Leftrightarrow5x-7\ge3x-6+x\)
\(\Leftrightarrow5x-7\ge4x-6\)
\(\Leftrightarrow5x-7-4x\ge4x-6-4x\)
\(\Leftrightarrow x-7\ge-6\)
\(\Leftrightarrow x-7+7\ge-6+7\)
\(\Leftrightarrow x\ge1\)
Vậy tập nghiệm của pt \(S\in\left\{1,\infty\right\}\)
trục số coi lại kt lớp 6 là được rồi
b,
\(3\left(4x+1\right)-2\left(5x+2\right)>8x-2\)
\(\Leftrightarrow12x+3-10x-4>8x-2\)
\(\Leftrightarrow2x-1>8x-2\)
\(\Leftrightarrow2x-1-8x>8x-2-8x\)
\(\Leftrightarrow-6x-1>-2\)
\(\Leftrightarrow-6x-1+1>-2+1\)
\(\Leftrightarrow-6x>-1\)
\(\Leftrightarrow\dfrac{-6x}{-6}< \dfrac{-1}{-6}\)
\(\Leftrightarrow x< \dfrac{1}{6}\)
c, \(1+x-\dfrac{x-3}{4}\ge\dfrac{x+1}{4}-\dfrac{x-2}{3}\)
\(\Leftrightarrow\dfrac{3}{4}.x+\dfrac{7}{4}\ge-\dfrac{1}{12}.x+\dfrac{11}{12}\)
\(\Leftrightarrow\dfrac{3}{4}.x+\dfrac{7}{4}+\dfrac{1}{12}.x\ge\dfrac{-1}{12}.x+\dfrac{11}{12}+\dfrac{1}{12}.x\)
\(\Leftrightarrow\dfrac{5}{6}.x+\dfrac{7}{4}\ge\dfrac{0}{12}.x+\dfrac{11}{12}\)
\(\Leftrightarrow\dfrac{5}{6}x+\dfrac{7}{4}-\dfrac{7}{4}\ge\dfrac{11}{12}-\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{5}{6}.x\ge-\dfrac{5}{6}\)
\(\Leftrightarrow x\ge\dfrac{-5}{6}.\dfrac{6}{5}\)
\(\Leftrightarrow x\ge-1\)
Vậy tập nghiệm của pt \(S\in\left\{-1,\infty\right\}\)
huhu giúp e với ạ e cần gấp ạ
