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Bài 3:
a: \(=35^{2018}\left(35-1\right)=35^{2018}\cdot34⋮17\)
b: \(=43^{2018}\left(1+43\right)=43^{2018}\cdot44⋮11\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Bài 1:
b:
x=9 nên x+1=10
\(M=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...-x\left(x+1\right)+x+1\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^2-x+x+1\)
=1
c: \(N=\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)+2^{10}\left(1+2+2^2+2^3+2^4\right)\)
\(=31\left(1+2^5+2^{10}\right)⋮31\)
Bài 1:
A = \(-\left(x-1\right)^2+5\)
Có: \(-\left(x-1\right)^2\)\(\le\)0
=> \(-\left(x-1\right)^2+5\)\(\le\)5
A min = 5 <=> x-1 = 0 <=> x =1
câu sau bạn làm tương tự
Bài 2:
a) n(n - 5) - (n-3)(n-2) = \(n^2-5n-n^2-5n+6=6\)chia hết cho 6
b) \(\left(n+2\right)^2-\left(n-2\right)^2=\left(n+2-n+2\right)\left(n+2+n-2\right)\)
= 4 . 2n = 8n chia hết cho 8
Bài 3: \(\left(ax-by\right)^2+\left(ay+bx\right)^2=a^2x^2-2axby+b^2y^2+a^2y^2+2axby+b^2x^2\)
= \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)
= \(\left(a^2+b^2\right)\left(x^2+y^2\right)\)
Đây là bất đẳng thức bu-nhi-a-cop-xki
Bài 1 :
a)Tìm giá trị nhỏ nhất của biểu thức
A = 2x2 - 4x + 8
\(=\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\sqrt{2}+\left(\sqrt{2}\right)^2+4\)
\(=\left(\sqrt{2}x+\sqrt{2}\right)^2+4\)
Ta có : \(\left(\sqrt{2}x+\sqrt{2}\right)^2\ge0\) \(\Rightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+4\ge4>0\)
=> A > 4
=> Amin = 4 \(\Leftrightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\sqrt{2}x+\sqrt{2}=0\)
\(\Leftrightarrow x=-1\)
Bài 1:
a) \(A=2x^2-4x+8\)
\(=2\left(x^2-2x+4\right)=2\left(x-2\right)^2\)
Xét \(2\left(x-2\right)^2\ge0\)
\(\Rightarrow Min_A=0\Leftrightarrow x=2\)
b) \(B=n^5-5n^3+4n\)
\(=n\left(n^4-5n^2+4\right)\)
\(=n\left(n^4-n^2-4n^2+4\right)\)
\(=n\left[n^2\left(n^2-1\right)-4\left(n^2-1\right)\right]\)
\(=n\left[\left(n^2-4\right)\left(n^2-1\right)\right]\)
\(=n\left[\left(n-2\right)\left(n+2\right)\left(n-1\right)\left(n+1\right)\right]\)
\(=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
Xét \(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\) là 5 số nguyên liên tiếp
\(\Rightarrow\)\(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮2;3;4;5\)
\(\Rightarrow\)\(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮120\)
\(\Rightarrow\)\(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮30\)
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
Câu 2:
\(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)\)
\(=n^2+5n-\left(n^2+2n-3n-6\right)\)
\(=n^2+5n-n^2+n+6=6n+6\)
\(=6\left(n+1\right)⋮6\)