a/ \(2n+12⋮n+2\)

Mà \(n+2⋮n+2\)

\(\Leftrightarrow\hept{\begin{cases}2n+12⋮n+2\\2n+4⋮n+2\end{cases}}\)

\(\Leftrightarrow8⋮n+2\)

\(\Leftrightarrow n+2\inƯ\left(8\right)\)

Suy ra :

+) n + 2 = 1 => n = -1 (loại)

+) n + 2 = 2 => n = 0

+) n + 2 = 4 => n = 2

+) n + 2 = 8 => n = 6

Vậy ......

b/ \(3n+5⋮n-2\)

Mà \(n-2⋮n-2\)

\(\Leftrightarrow\hept{\begin{cases}3n+5⋮n-2\\3n-6⋮n-2\end{cases}}\)

\(\Leftrightarrow11⋮n+2\)

\(\Leftrightarrow n+2\inƯ\left(11\right)\)

\(\Leftrightarrow\orbr{\begin{cases}n+2=1\\n+2=11\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}n=-1\left(loại\right)\\n=9\end{cases}}\)

Vậy ..

a/ \(\left(x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-3\\x^2=-1\left(loại\right)\end{cases}}\) 

Vậy ....

b/ \(\left(x+7\right)\left(x^2-36\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+7=0\\x^2-36=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-7\\x^2=36\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-7\\x=6or=-6\end{cases}}\)

Vậy ...

a. Δ' = b'² - ac = (m-1)² - (-2m-3) = m² - 2m + 1 + 2m + 3

= m² + 4 ≥ 4 > 0 ∀ m ∈ R

Vậy pt đã cho luôn có hai nghiệm x₁; x₂ phân biệt với mọi m thuộc R

b. Áp dụng Viet, ta có \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\x_1\cdot x_2=-2m-3\end{matrix}\right.\)

Theo đề ta có \(\left(4x_1+5\right)\left(4x_2+5\right)+19=0\)

⇔ \(16x_1x_2+20x_1+20x_2+25+19=0\)

⇔ \(16x_1x_2+20\left(x_1+x_2\right)+44=0\)

⇔ \(16\left(-2m-3\right)+20\left[-2\left(m-1\right)\right]+44=0\)

⇔ \(-32m-48-40m+40+44=0\)

⇔ \(-72m+36=0\Leftrightarrow m=\frac{1}{2}\)

Vậy với m = \(\frac{1}{2}\)thì pt đã cho có hai nghiệm x₁; x₂ thỏa mãn điều kiện \(\left(4x_1+5\right)\left(4x_2+5\right)+19=0\)

a) Ta có △\(=b^2-4ac=\left[2\left(m-1\right)\right]^2-4.1.\left(-2m-3\right)=4m^2-8m+4+8m+12=4m^2+16>0\)Vậy phương trình luôn có 2 nghiệm x₁,x₂ với mọi m thuộc R

b) Theo định lí Vi-ét với m\(\in R\), ta có

\(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=\frac{-2\left(m-1\right)}{1}=2-2m\\x_1x_2=\frac{c}{a}=\frac{-2m-3}{1}=-2m-3\end{matrix}\right.\)

Ta lại có \(\left(4x_1+5\right)\left(4x_2+5\right)+19=0\Leftrightarrow16x_1x_2+20x_1+20x_2+25+19=0\Leftrightarrow16x_1x_2+20\left(x_1+x_2\right)+44=0\Leftrightarrow16.\left(-2m-3\right)+20.\left(2-2m\right)+44=0\Leftrightarrow-32m-48+40-40m+44=0\Leftrightarrow-72m+36=0\Leftrightarrow m=\frac{1}{2}\)

Vậy m=\(\frac{1}{2}\) thì \(\left(4x_1+5\right)\left(4x_2+5\right)+19=0\)

4.a)n²(n+1)+2n(n+1)=(n+1)(n²+2n)=n(n+1)(n+2)

n,(n+1),(n+2) là ba số nguyên liên tiếp nên chia hết cho 2 và 3

\(\Rightarrow\)n(n+1)(n+2) chia hết cho 6

4 Chứng minh rằng:

a)\(n^2+\left(n+1\right)+2n\left(n+1\right)\) chia hết cho 6

Ta có:

\(n^2\left(n+1\right)+2n\left(n+1\right)\)

\(=n^3+3n^2+2n\)

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

\(=n\left(n+1\right)\left(n+2\right)\)

Ta thấy n , n+1 và n+2 là ba số tự nhiên liên tiếp

=> n(n+1) (n+2)\(⋮\)6

=> đpcm

b)\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8

Ta có:

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

\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)

\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)

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

\(=\left(2n-1\right).2\left(n-1\right).2n\)

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

=>\(4n\left(2n-1\right)\left(n-1\right)⋮4\left(1\right)\)

Mà(2n-1)(n-1)=(n+n-1)(n-1)

=>\(\left(2n-1\right)\left(n-1\right)⋮2\left(2\right)\)

Từ (1) và (2)=> Đpcm

c)\(\left(n+7\right)^2-\left(n-5\right)^2\) chia hết cho 24

Câu hỏi của Ngoc An Pham - Toán lớp 8 | Học trực tuyến

Chúc bạn học tốt!^^

a/Ta có: M(x)+N(x) = (2x⁵ - 4x³ + 2x² + 10x - 1) + (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10)

                              = 2x⁵ - 2x⁵ - 4x³ + 4x³ + 2x⁴ + 2x² + x² + 10x + x -1 - 10

                              = 2x⁴ + 3x² + 11x - 11

b/ Ta có: A(x) = N(x)-M(x) = (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10) - (2x⁵ - 4x³ + 2x² + 10x - 1)

                                         = -2x⁵ - 2x⁵ + 2x⁴ + 4x³ + 4x³ + x² - 2x² + x - 10x -10 + 1

                                         = -2x⁵ + 2x⁴ + 8x³ - x² - 9x -9

1/x(x-y)-4x+4y

=x(x-y)-4(x-y)

=(x-y)(x-4)

2/a)x^2-x=0

x(x-1)=0

<=>x=0 hoặc x-1=0

x =1

=>S={0;1}

b)(x+2)(x-3)-x-2=0

(x+2)(x-3)-(x+2)=0

(x+2)(x-3-1)=0

(x+2)(x-4)=0

<=>x+2=0 hoặc x-4=0

x =-2 x =4

=>{-2;4}

c)36^2-49=0

(6x-7)(6x+7)=0

<=>6x-7=0 hoặc 6x+7=0

6x =7 6x =-7

x =7/6 x =-7/6

=>{7/6;-7/6}

3/(n+7)^2-(n-5)^2

=(n+7-n+5)(n+7+n-5)

=12(2n+2)

=12*2(n+1)

=24(n+1) chia hết cho 24

=>(n+7)^2-(n-5)^2 chia hết cho 24.

Bài 1

d, \(x^2+2xy+y^2-2x-2y+1\)

\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)

\(\Rightarrow\left(x+y-1\right)^2\)

Bài 2:

a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\)

b,\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

c, \(4x^2-9=0\)

\(\Leftrightarrow4x^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)

d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)

\(\Leftrightarrow7x^2-16x+9=0\)

\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)

\(\Leftrightarrow x=\frac{16\pm2}{14}\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)

1.a)\(3x-3y+x^2-2xy+y^2\)

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

\(=\left(x-y\right)\left(3+x-y\right)\)

d)\(x^2+2xy+y^2-2x-2y+1\)

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

\(=\left(x+y+1\right)^2\)

2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)

\(\Leftrightarrow-5x-9=0\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)

b)\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)

c)\(4x^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)

d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)

\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)

3.Ta có:

Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)

\(\Rightarrow m+21=0\)

\(\Rightarrow m=-21\)

Vậy...!

ta chia A(x) cho B(x) được thương là: 4x-7 dư -m-21 nên để A(x) chia hết cho B(x) thì: -m-21 chia hết cho 2x-3 => m+21 chia hết cho 2x-3=>m+21=2xk-3k(k thuộc N)=> m=2xk-3k-21

Ta có: f(x) = x²+4x-5= x(x+4)-5

Để f(x)>0 thì x(x+4)>5

=> x\(\ge\)1

Để f(x)<0 thì x(x+4)<5

=> x < -4

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

Tớ làm cho bạn mà bạn toàn ko tick

a)a²(a+1)+2a(a+1)=(a²+2a)(a+1)=a(a+2)(a+1)

Ta có Ta có a(a+1)(a+2) là 3 số tự nhiên liên tiếp =>a(a+1)(a+2)⋮3 (1)

Mà a(a+1)\(⋮\)2 (2)

Từ (1)(2) suy ra a(a+1)(a+2)⋮6

=>a²(a+1)+2a(a+1)⋮6

b)a(2a-3)-2a(a+1)=2a²-3a-2a²-2a=-5a

Vì -5 chia hết 5

=>-5a chia hết 5

c)x²+2x+2=x²+2x+1+1=(x+1)²+1

Vì (x+1)²≥0

<=>(x+1)²+1>0

d)x²-x+1=\(x^2-\frac{2.1}{2}\)+\(\frac{1}{4}+\frac{3}{4}\)=\(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)(đpcm)

e)-x²+4x-5=-(x²-4x+5)=-(x²-4x+4)-1=-(x-2)²-1

Vì -(x-2)²≤0=>-(x-2)²-1<0(đpcm)

rồi nhé