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BA
29 tháng 7 2019
\(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)+4\left(3-x\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}}\)
BA
29 tháng 7 2019
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
JS
26 tháng 2 2020
Sửa đề: x2 + 13x + 41 --> x2 + 13x + 42
Giải:
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)
(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\)
\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)
\(\Leftrightarrow x^2+8x+7=12\)
⇔ x2−8x=5x2-8x=5
⇔ x2 −8x+(−4)2=5+(−4)2x2-8x+(-4)2=5+(-4)2
⇔ x2−8x+16=21x2-8x+16=21
⇔ (x−4)2=21(x-4)2=21
⇔ x=±√21+4x=±21+4
Vậy...
Chúc bạn học tốt@@
1 tháng 12 2022
a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)
\(=\dfrac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\)
b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)
c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)
d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)
10 tháng 7 2017
ĐK của A \(x\ne4\),ĐK của B \(\hept{\begin{cases}x\ne0\\x\ne5\end{cases}}\)
a, \(x^2-3x=0\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Với \(x=0\Rightarrow A=\frac{-5}{-4}=\frac{5}{4}\)
Với \(x=3\Rightarrow A=\frac{3-5}{3-4}=2\)
b. \(B=\frac{x+5}{2x}+\frac{x-6}{x-5}-\frac{2x^2-2x-50}{2x\left(x-5\right)}=\frac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-2x^2+2x+50}{2x\left(x-5\right)}\)
\(=\frac{x^2-10x+25}{2x\left(x-5\right)}=\frac{\left(x-5\right)^2}{2x\left(x-5\right)}=\frac{x-5}{2x}\)
c. \(P=\frac{A}{B}=\frac{x-5}{x-4}.\frac{2x}{x-5}=\frac{2x}{x-4}=\frac{2x-8}{x-4}+\frac{8}{x-4}=2+\frac{8}{x-4}\)
P nguyên \(\Leftrightarrow x-4\inƯ\left(8\right)\Rightarrow x-4\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow x\in\left\{-4;0;2;3;5;6;8;12\right\}\)
So sánh điều kiện ta thấy \(x\in\left\{-4;2;3;6;8;12\right\}\)thì P nguyên
B
5 tháng 3 2020
\(x^2-5x+6=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2-2x-3x+6=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x-2\right)-3\cdot\left(x+2\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-3\right)\cdot\left(x-2\right)=0\Rightarrow x\in\left(2,3\right)\)
\(x^2-7x+12=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2-3x-4x+12=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x-3\right)-4\left(x-3\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-4\right)\cdot\left(x-3\right)=0\Rightarrow x\in\left(3,4\right)\)
\(x^2+x-20=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2+5x-4x-20=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x+5\right)-4\cdot\left(x+5\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-4\right)\cdot\left(x+5\right)=0\Rightarrow x\in\left(4,-5\right)\)
câu 4 mk chịu
TN
5 tháng 3 2020
\(1.x^2-5x+6=0\\ x^2-2x-3x+6=0\\ \left(x^2-2x\right)+\left(-3x+6\right)=0\\ x\left(x-2\right)-3\left(x-2\right)=0\\ \left(x-2\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(2.x^2-7x+12=0\\ x^2-3x-4x+12=0\\ \left(x^2-4x\right)+\left(-3x+12\right)=0\\ x\left(x-4\right)-3\left(x-4\right)=0\\ \left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
\(3.x^2+x-20=0\\ x^2-4x+5x-20=0\\ \left(x^2-4x\right)+\left(5x-20\right)=0\\ x\left(x-4\right)+5\left(x-4\right)=0\\ \left(x-4\right)\left(x+5\right)=0\\ \left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
câu 4 mik nghĩ là đề sai
LA
21 tháng 6 2017
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
21 tháng 6 2017
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
7 tháng 11 2016
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
NN
10 tháng 11 2017
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
\(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x^2-5x=x-1\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-\frac{1}{5}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
\(2\left(x-7\right)-x^2+7x=0\)
\(2\left(x-7\right)-x\left(x-7\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)