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Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a) ta có :x2+2x+2=(x+1)2+1>0,với mọi x
x2+2x+3=(x+1)2+2>0,với mọi x
ĐKXĐ:x\(\in\)R.Đặt x2+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
=>6(a2-1)+6a2=7a2+7a<=>6a2-6+6a2=7a2+7a<=>12a2-7a2-7a-6=0
<=>5a2-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)
<=>a=2=>x2+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)
Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)
a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
\(=>\frac{1-x+x+1}{x+1}+2=\frac{1}{x+1}+2\)
\(=>\frac{2}{x+1}=\frac{1}{x+1}\)
\(=>2x+2=x+1\)
\(=>2x-x=1-2=-1\)
\(=>x=-1\)
vậy nghiệm của phương trình trên là {-1}
À quên ĐKXĐ của câu a là \(x\ne-1\)
Nên \(x\in\varnothing\)nhé :v
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>x=\(\dfrac{1}{5}\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
\(a.\Leftrightarrow x^2+x-6+2x^2+4x+2=x^2-6x+9-2x^2+4x\)
\(\Leftrightarrow4x^2+7x-13=0\)(pt vô nghiệm)
\(b.\Leftrightarrow x^3+3x^2+3x+1-x^2+2x+8=x^3-8+2x^2\)
\(\Leftrightarrow5x=-17\Rightarrow x=\frac{-17}{5}\)
Đặt \(t=x^2+2x+2\left(t\ge1\right)\)
\(c.\Leftrightarrow\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\)\(\Leftrightarrow\frac{t^2-1+t^2}{t^2+t}=\frac{7}{6}\)\(\Leftrightarrow12t^2-6=7t^2+7t\)
\(\Leftrightarrow5t^2-7t-6=0\Rightarrow\orbr{\begin{cases}t=2\left(tm\right)\\t=\frac{-3}{5}\left(l\right)\end{cases}}\)
\(\Rightarrow x^2+2x+2=2\Rightarrow x=-2\)