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\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)
Đặt: \(x^2+2x=t\)
khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)
\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)
Khi đó:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Đặt n^2 + 2n + 1= a, ta được:
(a - 1)(a + 1) +1= a^2 - 1 + 1= a^2=(n^2 + 2n +1)^2
=(n + 1)^4
2. Ta có: P = 2x2 + y2 - 4x - 4y + 10
P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4
P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)4 \(\forall\)x;y
=> P luôn dương với mọi biến x;y
3 Ta có:
(2n + 1)(n2 - 3n - 1) - 2n3 + 1
= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1
= -5n2 - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z
\(a,\left(2x-3\right)n-2n\left(n+2\right)\)
\(=n\left(2x-3-2n-4\right)\)
\(=-7n\)
Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM
\(b,n\left(2n-3\right)-2n\left(n+1\right)\)
\(=n\left(2n-3-2n-2\right)\)
\(=-5n⋮5\) (ĐPCM)
Rút gọn
\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)
\(=-76\)
\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)
\(=9\)
\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)
\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)
= -3
a) \(49-x^2+2xy-y^2\)
\(=49-\left(x^2-2xy+y^2\right)\)
\(=49-\left(x-y\right)^2\)
\(=\left(7-x+y\right)\left(7+x-y\right)\)
c) \(\frac{1}{36}a^2-\frac{1}{4}b^2\)
\(=\frac{1}{4}\left(\frac{1}{9}a^2-b^2\right)\)
\(=\frac{1}{4}\left(\frac{1}{3}a-b\right)\left(\frac{1}{3}a+b\right)\)
a) \(15x^ny^{2n}-3x^{n+1}\left(-y\right)^{2n}\)
\(=x^ny^{2n}\left(15-3x\right)\)
\(=3x^ny^{2n}\left(5-x\right)\)
b) \(4x^{2n}y^{n-1}+2\left(-x\right)^{2n+1}y^n\)
\(=4x^{2n}y^{n-1}-2x^{2n+1}y^n\)
\(=2x^{2n}y^{n-1}\left(2-xy\right)\)
a, 4y(x-1)-(1-x)
=(x-1)(4y+1)
b,3x(z+2)+5(-x-2)
=3x(z+2)-5(x+2)
=(z+2)(3x-5)
Đặt: \(x^2-6x+1=a;x^2+1=b\)
Khi đó đa thức này có dạng:
\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)
\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)
Thay lại a và b thì được:
\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)
\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)
\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)
Vậy ...
Bài 1:
\(x^5+x+1\)
\(=x^5-x^4+x^2+x^4-x^3+x+x^3-x^2+1\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Bài 2:
\(\frac{2n^2-3n+1}{2n+1}=\frac{n\left(2n+1\right)-4n+1}{2n+1}=\frac{n\left(2n+1\right)}{2n+1}-\frac{4n+1}{2n+1}=n-\frac{4n+1}{2n+1}\in Z\)
\(\Rightarrow4n+1⋮2n+1\)
\(\Rightarrow\frac{4n+1}{2n+1}=\frac{2\left(2n+1\right)-1}{2n+1}=\frac{2\left(2n+1\right)}{2n+1}-\frac{1}{2n+1}=2-\frac{1}{2n+1}\in Z\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow2n\in\left\{0;-2\right\}\)
\(\Rightarrow n\in\left\{0;-1\right\}\)