Phan tich da thuc thanh nhan tu x^2(1-x^2)-4-4x^2
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Sử dụng phép chia đa thức \(2n^2+5n-1\)cho n-1. Ta có được
\(2n^2+5n-1=\left(n-1\right)\left(2n+7\right)+6\)
Để \(2n^2+5n-1\)chia hết cho n-1 thì 6 phải chia hết cho n-1 => n-1 là ước của 6 ,
\(n-1\in U\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)và n-1 khác 0.
Bạn tự làm tiếp nhé!
Để 2n2 + 5n - 1 chia hết cho n - 1
=> 2n2 - 2n + 7n - 7 + 6 chia hết cho n - 1
2n.(n-1) + 7.(n-1) + 6 chia hết cho n - 1
(n-1).(2n+7) + 6 chia hết cho n - 1
mà (n-1).(2n+1) chia hết cho n - 1
=> 6 chia hết cho n - 1
=> n - 1 thuộc Ư(6)={1;-1;2;-2;3;-3;6;-6}
nếu n - 1 = 1 => n = 2 (TM)
...
bn tự xét tiếp nha!
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\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(A=\left(x+y\right).\left(x+y\right)^2\)
\(A=\left(x+y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(4x+26=-12\)
\(4x=-38\)
\(x=\frac{-19}{2}\)
x(3x+2) + (x+1)2 - (2x-5)(2x+5)= -12
(3x2+2x) + (x2+2x+1) - (4x2 - 25) = -12
3x2 + 2x + x2 + 2x + 1 - 4x2 +25 = -12
(3x2 + x2 - 4x2) + ( 2x+2x) + (1+25) = -12
0 + 4x + 26 = -12
4x = -12 - 26
4x = -38
x = -9.5
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a, \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
\(18m^2-36mn+18n^2-72p^2\)
\(=18\left(m^2-2mn+n^2-4p^2\right)\)
\(=18.\left[\left(m-n\right)^2-\left(2p\right)^2\right]\)
\(=18.\left(m-n-2p\right)\left(m-n+2p\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\)
\(\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\)
\(\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\)
\(-3\left(4x+5\right)=9\)
\(4x+5=-3\)
\(4x=-8\)
\(x=-2\)
b) \(x^2-2x-15=0\)
\(x^2-5x+3x-15=0\)
\(x\left(x-5\right)+3\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
c) \(2x^2+3x-5=0\)
\(2x^2-2x+5x-5=0\)
\(2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-7x+9\)
\(=x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{13}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2\)
\(=\left(x-\frac{7}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{7}{2}+\frac{\sqrt{13}}{2}\right)\)
\(=\left(x-\frac{7+\sqrt{13}}{2}\right)\left(x-\frac{7-\sqrt{13}}{2}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với a, b dương:
\(8^2=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)^2\ge\frac{4}{\sqrt{ab}}\)
\(\Rightarrow\frac{1}{\sqrt{ab}}\le\frac{64}{4}=16\)
max A=16 khi a=b=1/4