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a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\) (ĐK: \(x\ne0,x\ne-1\))
\(\Leftrightarrow\dfrac{360\left(x+1\right)}{x\left(x+1\right)}-\dfrac{400x}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow360\left(x+1\right)-400x=x\left(x+1\right)\)
\(\Leftrightarrow360x+360-400x=x^2+x\)
\(\Leftrightarrow-40x+360=x^2+x\)
\(\Leftrightarrow x^2+40x+x-360=0\)
\(\Leftrightarrow x^2+41x-360=0\)
\(\Rightarrow\Delta=41^2-4\cdot1\cdot\left(-360\right)=3121>0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-41+\sqrt{3121}}{2\cdot1}\approx7\left(tm\right)\\x_2=\dfrac{-41-\sqrt{3121}}{2\cdot1}\approx-48\left(tm\right)\end{matrix}\right.\)
\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\)
Điều kiện: \(x\ne0;x\ne-1\)
PT \(\Leftrightarrow\dfrac{360\left(x+1\right)-400x}{x\left(x+1\right)}=1\)
\(\Rightarrow-40x+360=x\left(x+1\right)\)
\(\Leftrightarrow-40x+360=x^2+x\)
\(\Leftrightarrow x^2+41x-360=0\)
\(\Leftrightarrow x^2+2.\dfrac{41}{2}.x+\dfrac{1681}{4}=\dfrac{3121}{4}\)
\(\Leftrightarrow\left(x+\dfrac{41}{2}\right)^2=\left(\dfrac{\sqrt{3121}}{2}\right)^2\)
\(\Leftrightarrow x+\dfrac{41}{2}=\dfrac{\sqrt{3121}}{2}\) hoặc \(x+\dfrac{41}{2}=-\dfrac{\sqrt{3121}}{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\) hoặc \(x=-\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\)
Vậy...
a, để pt trên là pt bậc nhất khi m khác 2
b, Ta có \(2x+5=x+7-1\Leftrightarrow x=1\)
Thay x = 1 vào pt (1) ta được
\(2\left(m-2\right)+3=m-5\Leftrightarrow2m-1=m-5\Leftrightarrow m=-4\)
Lưu Anh Đức: không được đặt dấu ''='' giữa 2 vế ở biểu thức cuối, 0x không thể bằng 1
x+\(\frac{1}{x}-1-x+\frac{1}{x}-1=\frac{4}{2x}-1\)
\(\Leftrightarrow\frac{2}{x}-2=\frac{4}{2x}-1\)
\(\Leftrightarrow\)\(\frac{4}{2x}-\frac{4}{2x}=-1+2\)
\(\Leftrightarrow0x=1\)
Vậy phương trình vô nghiệm
1.
Đặt \(x-2=t\ne0\Rightarrow x=t+2\)
\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)
\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)
2.
Đặt \(x-1=t\ne0\Rightarrow x=t+1\)
\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)
\(C_{max}=2\) khi \(t=3\) hay \(x=4\)
\(\Leftrightarrow4y^2=4x^4+4x^3+4x^2+4x+4\)
Ta có:
\(4x^4+4x^3+4x^2+4x+4=\left(2x^2+x\right)^2+2x^2+\left(x+2\right)^2>\left(2x^2+x\right)^2\)
\(4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+2\right)^2-5x^2\le\left(2x^2+x+2\right)^2\)
\(\Rightarrow\left(2x^2+x\right)^2< \left(2y\right)^2\le\left(2x^2+x+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}\left(2y\right)^2=\left(2x^2+x+1\right)^2\\\left(2y\right)^2=\left(2x^2+x+2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+1\right)^2\\4x^4+4x^3+4x^2+4x+4=\left(2x^2+x+2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x-3=0\\5x^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
- Với \(x=0\Rightarrow y^2=1\Rightarrow y=\pm1\)
- Với \(x=-1\Rightarrow y^2=1\Rightarrow y=\pm1\)
- Với \(x=3\Rightarrow y^2=121\Rightarrow y=\pm11\)
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tick nhé