Tìm x nguyên dương để 1+2x+3x+4x chia hết cho 5
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a) \(\left(x^2+y^2-36\right)^2-4x^2y^2\)
\(=\left(x^2+y^2-36\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+y^2+2xy-36\right)\left(x^2+y^2-2xy-36\right)\)
\(=\left[\left(x+y\right)^2-36\right]\left[\left(x-y\right)^2-36\right]\)
\(=\left(x+y+6\right)\left(x+y-6\right)\left(x-y+6\right)\left(x-y-6\right)\)
b) \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-3\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x-3\right)\left(x-2\right)\left(x+1\right)\)
1) a) (x2 + y2 - 36)2 - 4x2y2
= (x2 + y2 - 36 - 2xy)(x2 + y2 - 36 + 2xy)
= [(x - y)2 - 36][(x + y)2 - 36]
= (x - y - 6)(x - y + 6)(x + y + 6)(x + y - 6)
b) (x2 + x)2 - 5(x2 + x) + 6
= (x2 + x)2 - 2(x2 + x) - 3(x2 + x) + 6
= (x2 + x)(x2 + x - 2) - 3(x2 + x - 2)
= (x2 + x - 3)(x2 + 2x - x - 2)
= (x2 + x - 3)(x - 1)(x + 2)
2) Đặt tính là đc
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(P=x^3+y^3+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y-2\right)\)
\(=2^3=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+x+1=x\left(x+1\right)+1\)
Vì \(x\inℤ\)\(\Rightarrow x\left(x+1\right)⋮x+1\)\(\Rightarrow\)Để \(x^2+x+1⋮x+1\)thì \(1⋮x+1\)
\(\Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{-2;0\right\}\)
Vậy \(x\in\left\{-2;0\right\}\)
b) \(3x-8=3x-12+4=3\left(x-4\right)+4\)
Vì \(3\left(x-4\right)⋮x-4\)\(\Rightarrow\)Để \(3x-8⋮x-4\)thì \(4⋮x-4\)
\(\Rightarrow x-4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Lập bảng giá trị ta có:
\(x-4\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) |
\(x\) | \(0\) | \(2\) | \(3\) | \(5\) | \(6\) | \(8\) |
Vậy \(x\in\left\{0;2;3;5;6;8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng định lý Bezout: f(x) chia hết cho ax + b \(\Leftrightarrow f\left(\frac{-b}{a}\right)=0\)
Đặt \(g\left(x\right)=4x^4+2x^3+3x^2-4x+5+m\)
Để đa thức \(g\left(x\right)=4x^4+2x^3+3x^2-4x+5+m\)chia hết cho nhị thức 2x + 3 thì :
\(g\left(\frac{-3}{2}\right)=4.\left(\frac{-3}{2}\right)^4+2.\left(\frac{-3}{2}\right)^3+3.\left(\frac{-3}{2}\right)^2-4.\frac{-3}{2}+5+m=0\)
\(\Leftrightarrow\frac{81}{4}-\frac{27}{4}+\frac{27}{4}+6+5+m=0\)
\(\Leftrightarrow\frac{81}{4}-11+m=0\)
\(\Leftrightarrow\frac{37}{4}+m=0\)
\(\Leftrightarrow m=\frac{-37}{4}\)
Vậy \(m=\frac{-37}{4}\)thì \(4x^4+2x^3+3x^2-4x+5+m\)chia hết cho 2x + 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(x-3\right)+5⋮x-3\Rightarrow x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
x-3 | 1 | -1 | 5 | -5 |
x | 4 | 2 | 8 | -2(ktm) |
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x-1⋮x-3\)
\(=>2.\left(x-3\right)+5⋮x-3\)
Do \(2.\left(x-3\right)⋮x-3\)
\(=>5⋮x-3\)
\(=>x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(=>x\in\left\{-2;2;4;8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(2x-1=2\left(x-3\right)+5\)
để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)
ta có bảng sau :
x-3 | -5 | 5 | -1 | 1 |
x | -2 | 2 | 4 | 8 |
\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)
\(do2.\left(x-3\right)⋮x-3\)
\(\Leftrightarrow5⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)