Bài 8:

a) \(x^2-25x=0\)

\(\Leftrightarrow x\left(x-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)

b) \(7x\left(x-3\right)-5\left(3-x\right)=0\)

\(\Leftrightarrow7x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{7}\end{matrix}\right.\)

c) \(7x\left(x+4\right)-7x-28=0\)

\(\Leftrightarrow7x\left(x+4\right)-7\left(x+4\right)=0\)

\(\Leftrightarrow7\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

d) \(\left(2x^2+5x\right)\left(x^2-x\right)+\left(2x^2+10\right)\left(x-x^2\right)=0\)

\(\Leftrightarrow\left(2x^2+5x\right)\left(x^2-x\right)-\left(2x^2+10\right)\left(x^2-x\right)=0\)

\(\Leftrightarrow\left(x^2-x\right)\left(2x^2+5x-2x^2-10\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\5x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

7:

a: \(4a^3b-12a^2b^2+8ab^3\)

\(=4ab\cdot a^2-4ab\cdot3ab+4ab\cdot2b^2\)

\(=4ab\left(a^2-3ab+2b^2\right)\)

\(=4ab\left(a^2-ab-2ab+2b^2\right)\)

\(=4ab\left[a\left(a-b\right)-2b\left(a-b\right)\right]\)

\(=4ab\left(a-b\right)\left(a-2b\right)\)

b: \(\dfrac{5}{2}x^4+\dfrac{3}{4}x^3-\dfrac{1}{5}x^2\)

\(=x^2\cdot\dfrac{5}{2}x^2+x^2\cdot\dfrac{3}{4}x-x^2\cdot\dfrac{1}{5}\)

\(=x^2\left(\dfrac{5}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{5}\right)\)

c: \(3x^2\left(x+9\right)-2\left(x+9\right)\)

\(=\left(x+9\right)\cdot3x^2-\left(x+9\right)\cdot2\)

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

d: \(3x^2\left(7x+y\right)-5x\left(7x+y\right)+14x+2y\)

\(=\left(7x+y\right)\left(3x^2-5x\right)+2\left(7x+y\right)\)

\(=\left(7x+y\right)\left(3x^2-5x+2\right)\)

\(=\left(7x+y\right)\left(3x^2-3x-2x+2\right)\)

\(=\left(7x+y\right)\left[3x\left(x-1\right)-2\left(x-1\right)\right]\)

\(=\left(7x+y\right)\left(x-1\right)\left(3x-2\right)\)

\(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

\(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(b+1\right)}=\dfrac{b}{a}\)

d) Để phân thức \(\dfrac{4x^3+4x^2}{x^2-1}\) có nghĩa thì: \(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

Khi đó: \(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

e) Để phân thức \(\dfrac{b^2+b}{a+ab}\) có nghĩa thì: \(a+ab\ne0\Leftrightarrow a\ne-ab\)

Khi đó: \(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(1+b\right)}=\dfrac{b}{a}\)

Giai

Đầu tiên bạn tính số số hạng của tổng là 672

Vì 672 : 4 = 163 nên ta có :

S = ( 3-6-9+12 ) + ( 15-18-21+24) + .......+ (2007-2010-2013+2016)

S = 0 + 0 + ..... + 0 ( 163 số )

S = 0 . 163 = 0

tick minh nha

4:

a: \(=5x^2y^2\cdot\dfrac{7}{10}x^4y\cdot40x^2z^3\)

\(=\left(5\cdot\dfrac{7}{10}\cdot40\right)\left(x^2y^2\cdot x^4y\cdot x^2z^3\right)\)

\(=140x^8y^3z^3\)

Bậc là 8+3+3=14

b: \(=-\dfrac{1}{2}\cdot ab\cdot\dfrac{-4}{3}\cdot a^2bc\cdot5c^2b^3\)

\(=\left(\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot5\right)\cdot ab\cdot a^2b\cdot c\cdot c^2b^3\)

\(=\dfrac{10}{3}a^3b^5c^3\)

Bậc là 11

c: \(=-1.2\cdot ab\cdot100a^4b^2\cdot c^3\cdot\left(-1.5\right)\cdot a^2c\)

\(=1.8\cdot100\cdot ab\cdot a^4b^2c^3\cdot a^2c\)

\(=180a^7b^3c^4\)

Bậc là 14

Bài 12:

a) \(4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot1+1^2=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(x^2-2x=-1\)

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

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

c) \(-x^2+10x=25\)

\(\Leftrightarrow-x^2+10x-25=0\)

\(\Leftrightarrow-\left(x^2-10x+25\right)=0\)

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

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

d) \(9=12x-4x^2\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Bài 2: 

a: Xét ΔBHA vuông tại H và ΔBHD vuông tại H có 

BH chung

HA=HD

Do đó: ΔBHA=ΔBHD

b: Ta có: ΔBHA=ΔBHD

nên \(\widehat{ABH}=\widehat{DBH}\)

hay BH là tia phân giác của góc ABD

Gọi độ dài AB và vận tốc dự kiến lần lượt là x,y

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{10}{3}\\\dfrac{x}{y+5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-10y=0\\x-3y=15\end{matrix}\right.\)

=>x=150 và y=45

Bài 6:

Xét tứ giác AECF có 

AE//CF

AE=CF

Do đó: AECF là hình bình hành