Câu 2: Tìm x:

b) \(10x+25+x^2=0\)

<=> \(x^2+2.5x+5^2=0\)

<=> \(\left(x+5\right)^2=0\)

=> \(x+5=0\)

<=> \(x=-5\)

Vậy x= \(-5\)

a) \(2\left(x^2+3\right)-2x\left(x+3\right)=0\)

<=> \(2x^2+6-2x^2-6x=0\)

<=> \(6-6x=0\)

<=> \(6\left(1-x\right)=0\)

=> \(1-x=0\)

<=> \(x=1\)

Vậy x = 1

Câu 1.2: Rút gọn biểu thức:

a) \(2x\left(x+3\right)-x\left(2x-1\right)\)

\(=2x^2+6x-2x^2+x\)

\(=7x\)

b) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)

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

\(=x^3-\left(2y\right)^3\)

\(=x^3-8y^3\)

c) \(\left(x+2\right)^2+\left(x-2\right)^2\)

\(=x^2+4x+4+x^2-4x+4\)

\(=2x^2+8\)

\(=2\left(x^2+4\right)\)

d) \(\left(4y+1\right)^2-4\left(2y+3\right)\left(2y-3\right)\)

\(=\left(4y+1\right)^2-4\left(\left(2y\right)^2-3^2\right)\)

\(=\left(4y+1\right)^2-4\left(4y^2-9\right)\)

\(=16y^2+8y+1-16y^2+36\)

\(=8y+37\)

A= \(n^3-4n^2+4n-1\)

\(=\left(n^3-1\right)-\left(4n^2-4n\right)\)

\(=\left(n-1\right)\left(n^2+n+1\right)-4n\left(n-1\right)\)

\(=\left(n-1\right)\left(n^2+n+1-4n\right)\)

\(=\left(n-1\right)\left(n^2-3n+1\right)\)

=> A= \(n^3-4n^2+4n-1\) là số nguyên tố khi

\(n-1=1\) hoặc \(n^2-3n+1=1\) ;

với n là số tự nhiên:

* Với \(n-1=1\) <=> n=2 => A = \(-1\) (loại)

* Với \(n^2-3n+1=1\)

<=> \(n^2-3n=0\)

<=> \(n\left(n-3\right)=0\)

1/ n=0 => A = \(-1\) (loại)

2/ n - 3 =0 <=> n = 3 => A = 2 (thoã mãn)

Vậy A = \(n^3-4n^2+4n-1\) là số nguyên tố khi n=3

\(\dfrac{2x^3-4x^2}{x^2+8x+16}.\dfrac{3x+12}{4x-x^3}\)

(ĐKXĐ: x ≠ \(-4\) ; x ≠ 0; x ≠ 2 ; x ≠ \(-2\) )

\(\dfrac{2x^3-4x^2}{x^2+8x+16}.\dfrac{3x+12}{4x-x^3}\)

\(=\dfrac{2x^2\left(x-2\right)}{\left(x+4\right)^2}.\dfrac{3\left(x+4\right)}{x\left(4-x^2\right)}\)

\(=\dfrac{2x^2\left(x-2\right)}{\left(x+4\right)^2}.\dfrac{3\left(x+4\right)}{x\left(2-x\right)\left(2+x\right)}\)

\(=\dfrac{6x^2\left(x-2\right)\left(x+4\right)}{x\left(x+4\right)^2\left(2-x\right)\left(2+x\right)}\)

\(=\dfrac{-6x^2\left(2-x\right)\left(x+4\right)}{x\left(x+4\right)^2\left(2-x\right)\left(2+x\right)}\)

\(=\dfrac{-6x}{\left(x+4\right)\left(x+2\right)}\)

\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\);

( ĐKXĐ: x ≠ 0; x ≠ \(-1\); x ≠ 4; x ≠ \(-2\) )

\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\)

\(=\dfrac{x+1}{x^2-4x+2x-8}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{x+1}{x\left(x-4\right)+2\left(x-4\right)}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{x+1}{\left(x-4\right)\left(x+2\right)}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{4-x}{x\left(x-4\right)\left(x+2\right)}\)

=\(\dfrac{-\left(x-4\right)}{x\left(x-4\right)\left(x+2\right)}\)

\(=\dfrac{-1}{x\left(x+2\right)}\)

1 người đào được số mét mương là:

      36/10=3,6(m)

25 người đào được số mét mương là:

       3,6*25=90(m)

             Đáp số: 90 m

\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\) ;

(ĐKXĐ: x ≠ 0; x ≠ \(-1\); x ≠ 4; x ≠ \(-2\))

\(\dfrac{x+1}{x^2-2x-8}.\dfrac{4-x}{x^2+x}\)

\(=\dfrac{x+1}{x^2-4x+2x-8}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{x+1}{x\left(x-4\right)+2\left(x-4\right)}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{x+1}{\left(x-4\right)\left(x+2\right)}.\dfrac{4-x}{x\left(x+1\right)}\)

\(=\dfrac{4-x}{x\left(x-4\right)\left(x+2\right)}\)

\(=\dfrac{-\left(x-4\right)}{-x\left(x-4\right)\left(x+2\right)}\)

\(=\dfrac{-1}{-x\left(x+2\right)}\)

\(=\dfrac{1}{x\left(x+2\right)}\)

\(CM:\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)=2x+1\)

(ĐKXĐ: x ≠ 0; x ≠ 1 )

*Biến đổi vế trái :

\(\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)\)

\(=\dfrac{x^3-1}{x}\left(\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)

\(=\dfrac{x^3-1}{x}.\dfrac{\left(x+1\right)\left(x-1\right)+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^3-1}{x}.\dfrac{x^2-1+x^2+x+1}{x^3-1}\)

\(=\dfrac{2x^2+x}{x}=\dfrac{x\left(2x+1\right)}{x}\)

\(=2x+1\)

Vậy:

\(\left(x^2-\dfrac{1}{x}\right)\left(\dfrac{x+1}{x^2+x+1}+\dfrac{1}{x-1}\right)=2x+1\)

Phương trình  2 z 2 + 4 z + 5 = 0 có các nghiệm là

A.  2 ± i 6 2

B.  1 2 ± i 6 2

C.  - 1 ± i 6 2

D.  1 ± i 6 2