b. Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\
\Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\) \(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\) Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\) \(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \) Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\) a ) \(\left|x+5\right|=3x+1\) ( 1 ) + ) \(x+5=x+5.\) Khi \(x\ge-5\) \(\left(1\right)\Leftrightarrow x+5=3x+1\) \(\Leftrightarrow-2x=-4\Leftrightarrow x=2\) ( TM ) + ) \(x+5=-x-5.\) Khi \(x< -5\) \(\left(1\right)\Leftrightarrow-x-5=3x+1\) \(\Leftrightarrow-4x=6\) \(\Leftrightarrow x=-\dfrac{3}{2}\)( KTM ) Vậy .......... b ) \(\dfrac{3\left(x-1\right)}{4}+1\ge\dfrac{x+2}{3}\) \(\Leftrightarrow9x-9+12\ge4x+8\) \(\Leftrightarrow5x\ge5\) \(\Leftrightarrow x\ge1\) Vậy ........... c ) \(\dfrac{x-2}{x+1}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(1\right)\) ĐKXĐ : \(x\ne2;x\ne-2.\) \(\left(1\right)\Leftrightarrow\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\) \(\Rightarrow\left(x-2\right)^2-3\left(x+2\right)=2x-22\) \(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\) \(\Leftrightarrow x^2-9x+20=0\) \(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(TMĐKXĐ\right)\) Vậy ......... \(\Leftrightarrow\) a) 4x -8 ≥ 3(3x-1)-2x +1 ⇒4x -8 ≥7x -2 ⇒4x -7x ≥ -2 +8 ⇒-3x ≥ 6 ⇒x≤-2 Vậy bpt có nghiệm là:{x|x≤-2} b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4 ⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4 ⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16 ⇔ -3x ≤ -6 ⇔ x≥ 2 Vậy bpt có tập nghiệm là: {x|x≥2} a) \(4x-10< 0\\
4x< 10\\
x< \dfrac{10}{4}=\dfrac{5}{2}\) b) \(2x+x+12\ge0\\
3x\ge-12\\
x\ge-\dfrac{12}{3}=-4\) c) \(x-5\ge3-x\\
2x\ge8\\
x\ge4\) d) \(7-3x>9-x\\
-2>2x\\
x< -1\) đ) \(2x-\left(3-5x\right)\le4\left(x+3\right)\\
2x-3+5x\le4x+12\\
3x\le15\\
x\le5\) e) \(3x-6+x< 9-x\\
5x< 15\\
x< 3\) f) \(2t-3+5t\ge4t+12\\
3t\ge15\\
t\ge5\) g) \(3y-2\le2y-3\\
y\le-1\) h) \(3-4x+24+6x\ge x+27+3x\\
0\ge2x\\
0\ge x\) i) \(5-\left(6-x\right)\le4\left(3-2x\right)\\
5-6+x\le12-8x\\
\\
9x\le13\\
x\le\dfrac{13}{9}\) k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\
10x-15-20x+28\ge19-2x-22\\
13-10x\ge-2x-3\\
-8x\ge-16\\
x\le\dfrac{-16}{-8}=2\) l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\
\dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\
\Rightarrow40x-100-90x+30< 36-12x-30x+15\\
130-50x< 51-42x\\
92x< -79\\
x< -\dfrac{79}{92}\) m) \(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\
\dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\
\Rightarrow10x-3+2x>7x-5+2x\\
12x-3>9x-5\\
3x>-2\\
x>-\dfrac{2}{3}\) n) \(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\
\dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\
\Rightarrow28x-8-24x< 60-3x+6\\
4x-8< -3x+66\\
7x< 74\\
x< \dfrac{74}{7}\) a) \(4x-10< 0\) \(\Leftrightarrow4x< 10\) \(\Leftrightarrow x< \dfrac{5}{2}\) b) ??? c) \(x-5\ge3-x\) \(\Leftrightarrow2x-5\ge3\) \(\Leftrightarrow2x\ge8\) \(\Leftrightarrow x\ge4\) d) \(7-3x>9-x\) \(\Leftrightarrow7-2x>9\) \(\Leftrightarrow-2x>2\) \(\Leftrightarrow x< -1\) đ) ??? e) \(3x-6+x< 9-x\) \(\Leftrightarrow4x-6< 9-x\) \(\Leftrightarrow5x-6< 9\) \(\Leftrightarrow5x< 15\) \(\Leftrightarrow x< 3\) f) ??? g) ??? h) \(3-4x+24+6x\ge x+27+3x\) \(\Leftrightarrow2x+27\ge4x+27\) \(\Leftrightarrow-2x\ge0\) \(\Leftrightarrow x\le0\) i) \(5-\left(6-x\right)\le4\left(3-2x\right)\) \(\Leftrightarrow5-6+x\le12-8x\) \(\Leftrightarrow x-1\le12-8x\) \(\Leftrightarrow9x-1\le12\) \(\Leftrightarrow9x\le13\) \(\Leftrightarrow x\le\dfrac{13}{9}\) k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\) \(\Leftrightarrow10x-15-20x+28\ge19-2x-22\) \(\Leftrightarrow-10x+23\ge-3-2x\) \(\Leftrightarrow-8x+13\ge-3\) \(\Leftrightarrow-8x\ge-16\) \(\Leftrightarrow x\ge2\) l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\) \(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\) \(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\) \(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\) \(\Leftrightarrow x>-\dfrac{121}{8}\) m, n) làm tương tự: đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\) 2.a) \(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\) \(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\) \(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\) \(\Leftrightarrow-3x>6\) \(\Leftrightarrow3>\dfrac{6}{-3}\) \(\Leftrightarrow x< -2\) Vậy nghiệm của bpt \(S=\left\{-2\right\}\) 2.b) \(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\) \(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\) \(\Leftrightarrow4x+4-12\ge3x-6\) \(\Leftrightarrow4x-3x\ge-6-4+12\) \(\Leftrightarrow x\ge2\) vậy nghiệm của bpt x\(\ge\)2
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