Ôn tập toán 8

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

a) \(x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\\x=-3\end{array}\right.\)

b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)-\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(-2x-7-2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(-4x-12\right)=0\)

\(\Leftrightarrow-4\left(2x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-5=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

c) \(x^2-2x-3=0\)

\(\Leftrightarrow x^2+x-3x-3=0\)

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\x-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=3\end{array}\right.\)

c) \(2x^2+7x=4\)

\(\Leftrightarrow2x^2+7x-4=0\)

\(\Leftrightarrow2x^2+8x-x-4=0\)

\(\Leftrightarrow2x\left(x+4\right)-\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+4=0\\2x-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)

e) \(3x^2-7x+2=0\)

\(\Leftrightarrow3x^2-x-6x+2=0\)

\(\Leftrightarrow x\left(3x-1\right)-2\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=1\end{array}\right.\)

a) A = (2x - 1)(x - 3) 

=2x²-6x-x+3

=2x²-7x+3

\(=2\left(x^2-\frac{7x}{2}+\frac{3}{2}\right)\)

\(=2\left(x^2-\frac{7x}{2}+\frac{49}{16}-\frac{50}{16}\right)\)

\(=2\left(x-\frac{7}{4}\right)^2-\frac{25}{8}\ge0-\frac{25}{8}=-\frac{25}{8}\)

Dấu = khi \(x=\frac{7}{4}\)

Vậy MinA\(=-\frac{25}{8}\) khi \(x=\frac{7}{4}\)

b) B = (1 - 2x)(x - 3) 

=x-3-2x²+6x

=-2x²+7x-3

\(=-2\left(x^2-\frac{7x}{2}+\frac{3}{2}\right)\)

\(=-2\left(x^2-\frac{7x}{2}+\frac{49}{16}-\frac{50}{16}\right)\)

\(=\frac{25}{8}-2\left(x-\frac{7}{4}\right)^2\le\frac{25}{8}-0=\frac{25}{8}\)

Dấu = khi \(x=\frac{7}{4}\)

Vậy MaxA\(=\frac{25}{8}\) khi \(x=\frac{7}{4}\)

Gọi tia đối tia OB là tia Ox và cắt AC lại K.

Ta có: góc BOC là góc ngoài ▲ KOC tại đỉnh O

→ góc BOC = góc ACO + góc OKC

MK : góc OKC là góc ngoài ▲ ABK

→ góc OKC = góc A + góc ABO

→ góc BOC = góc A + góc ABO + góc ACO

CHÚC BẠN HỌC TỐT !!!

\(x^7+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left[\left(x^3\right)^2-1^2\right]+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left[x^2\left(x^3+1\right)-x\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5+x^2-x^4-x+1\right)\)

Bài 1 A=xyz+xz-zy-z+xy+x-y-1

thay các gtri x=-9, y=-21 và z=-31 vào là đc

=> A=-7680

Bài 2:a) n³ + 3n² + 2n = n²(n + 1) + 2n(n + 1) = n(n + 1)(n + 2)
số chia hết cho 6 là số chia hết cho 2 và 3
mà (n + 1) chia hết cho 2 và 3 với mọi số nguyên n
(n + 2) chia hết cho 2 và 3 với mọi số nguyên n
=>n³ + 3n² + 2n luôn chia hết cho 6 với mọi số nguyên n

b) 49ⁿ+77ⁿ-29ⁿ-1

=\(49^n-1+77^n-29^n\)

=\(\left(49-1\right)\left(49^{n-1}+49^{n-2}+...+49+1\right)+\left(77-29\right)\left(79^{n-1}+..+29^n\right)\)

=48(\(49^{n-1}+...+1+77^{n-1}+...+29^{n-1}\))

=> tích trên chia hết 48

c) 35x-14y+2⁹-1=7(5x-2y)+7.73

=7(5x-2y+73) tích trên chia hết cho 7

=. ĐPCM

Ta coˊ :xy+x+1x​+yz+y+1y​+xz+z+1z​

=���+�+1+�����+��+�+����2��+���+��=xy+x+1x​+xyz+xy+xxy​+x2yz+xyz+xyxyz​

=���+�+1+����+�+1+1��+�+1(Vıˋ ���=1)=xy+x+1x​+xy+x+1xy​+xy+x+11​(Vıˋ xyz=1)

=�+��+1��+�+1=xy+x+1x+xy+1​

$=1$

\(x^5+x+1\)

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

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

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

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

\(x^5+x+1\)

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

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

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

\(x^2-3x+4\)

\(=x^2-2.x.\frac{3}{2}+\left(\frac{2}{3}\right)^2+\frac{7}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0;\frac{7}{4}>0\)

=> Đa thưc vô nghiệm

\(x^2-3x+4=x^2-2.x.\frac{3}{2}+\frac{9}{4}+4-\frac{9}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\) ( vô nghiệm )

Vậy \(x^2-3x+4\) vô nghiệm

\(x^2-3x+4=x^2-2\cdot\frac{3}{2}\cdot x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

Vậy đa thức trên k có nghiệm