Bài 1:

Giải:

Đặt \(a=5k+4\)

Ta có: \(a^2=\left(5k+4\right)^2=25k^2+40k+16\)

\(=25k^2+40k+15+1\)

\(=5\left(5k^2+8k+3\right)+1\)

\(\Rightarrow a^2\) chia 5 dư 1

Vậy...

Bài 2:

a, Thay x = 87, y = 13 vào A có:

\(A=87^2-13^2=\left(87+13\right)\left(87-13\right)\)

\(=100.74=7400\)

Vậy A = 7400

b, \(B=x^3-3x^2+3x-1\)

\(=\left(x-1\right)^3\)

Thay x = 101 \(\Rightarrow B=100^3=1000000\)

Vậy B = 1000000

c, \(C=x^3+9x^2+27x+27=\left(x+3\right)^3\)

Thay x = 79 \(\Rightarrow C=82^3\)

Vậy \(C=82^3\)

\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{24}\)

Vậy x = \(\dfrac{-10}{24}\)

\(x^2-2xy-4z^2+y^2\)

\(=\left(x-y\right)^2-4z^2\)

\(=\left(x-y-2z\right)\left(x-y+2z\right)\)

a, \(A=4-x^2+2x=-\left(x^2-2x-4\right)\)

\(=-\left(x^2-2x+1-5\right)\)

\(=-\left(x-1\right)^2+5\le5\)

Dấu " = " khi \(-\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy \(MAX_A=5\) khi x = 1

\(S=\dfrac{1}{1.4.7}+\dfrac{1}{4.7.10}+...+\dfrac{1}{22.25.28}\)

\(=\dfrac{1}{6}\left(\dfrac{6}{1.4.7}+\dfrac{6}{4.7.10}+...+\dfrac{6}{22.25.28}\right)\)

\(=\dfrac{1}{6}\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+...+\dfrac{1}{22.25}-\dfrac{1}{25.28}\right)\)

\(=\dfrac{1}{6}\left(\dfrac{1}{4}-\dfrac{1}{25.28}\right)\)

\(=\dfrac{1}{24}-\dfrac{1}{6.25.28}\)

Vậy...

\(10x^2-19x=33\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2-30x+11x-33=0\)

\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)

\(\Leftrightarrow\left(10x+11\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+11=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{10}\\x=3\end{matrix}\right.\)

Vậy \(x=\dfrac{-11}{10}\) hoặc x = 3

\(10n^2+n-10=10n^2-10n+11n-11+1=10n\left(n-1\right)+11\left(n-1\right)+1\)

\(Để:10n^2+n-10\)chia hết cho n-1 thì 1 chia hết cho n-1 => n-1 =1 => n =2 hoặc n-1 =-1 => n =0

Vậy n = 0 ; 2

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)

\(=\left(x^2y+x^2z+xz^2+xyz\right)+\left(xy^2+y^2z+yz^2+xyz\right)\)

\(=x\left(xy+xz+z^2+yz\right)+y\left(xy+yz+z^2+xz\right)\)

\(=\left(x+y\right)\left(xy+yz+xz+z^2\right)\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

a, \(x^2-7x+12=x^2-4x-3x+12\)

\(=x\left(x-4\right)-3\left(x-4\right)=\left(x-3\right)\left(x-4\right)\)

b, \(x^2-5x-14=x^2+2x-7x-14\)

\(=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\)

c, \(4x^2-3x-1=4x^2-4x+x-1\)

\(=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)

\(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x-y-1\right)\left(x+y\right)\)