over

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was

couldn't

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takes

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go around 

go through

wind

how long

for

do

beautifully

excitedly

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arrangements

peacefully

excited

different

a,\(x^2-7x+6=x^2-x-6x+6\)

                       \(=x\left(x-1\right)-6\left(x-1\right)\)

                        \(=\left(x-6\right)\left(x-1\right)\)

a) x²-7x+6=(x²-x)-(6x-6)=x(x-1)-6(x-1)=(x-6)(x-1)

b) x²-6x+3=(x²-6x+9)-6=(x-3)²-\(\sqrt{6^2}\)=(x-3-\(\sqrt{6}\))(x-3+\(\sqrt{6}\))

c) x²-4x+3=(x²-x)-(3x-3)=x(x-1)-3(x-1)=(x-3)(x-1)

d) 3x²-5x+2=(3x²-3x)-(2x-2)=3x(x-1)-2(x-1)=(3x-2)(x-1)

e) 7x²-x-6=(7x²-7x)+(6x-6)=7x(x-1)+6(x-1)=(7x+6)(x-1)

f) 3x²-5x-8=(3x²+3x)-(8x+8)=3x(x+1)-8(x+1)=(3x-8)(x+1)

g) x²-6x+5=(x²-x)-(5x-5)=x(x-1)-5(x-1)=(x-5)(x-1)

h) x²-2x-3=(x²-2x+1)-4=(x-1)²-2²=(x-1-2)(x-1+2)=(x-3)(x+1)

i) x²-x-12=(x²+3x)-(4x+12)=x(x+3)-4(x+3)=(x-4)(x+3)

Bài 4:

a: a\(\perp\)c

b\(\perp\)c

Do đó: a//b

lười học thế

suốt ngày chép mạng

d, gt=> (x-3)³-(x-3)=0

<=>(x-3)(x²-6x+9-1)=0 <=>x=3 hoặc x=4 hoặc x=2

Bài cuối mình không thấy rõ đề nhưng mình đoán là thế này bạn nhé.

a.

$ax-ay+bx-by=(ax-ay)+(bx-by)=a(x-y)+b(x-y)$

$=(a+b)(x-y)$

b. Trùng phần a

c. 

$x^2+xy+ax+ay=(x^2+xy)+(ax+ay)=x(x+y)+a(x+y)$

$=(x+a)(x+y)$

d.

$x^2y+xy^2-x-y=(x^2y+xy^2)-(x+y)$

$=xy(x+y)-(x+y)=(xy-1)(x+y)$

e.

$x^2y-xy^2-5x+5y=(x^2y-xy^2)-(5x-5y)$

$=xy(x-y)-5(x-y)=(x-y)(xy-5)$

f. Biểu thức không phân tích được thành nhân tử

g.

$x^2y-4x+xy^2-4y$

$=(x^2y+xy^2)-(4x+4y)=xy(x+y)-4(x+y)=(x+y)(xy-4)$
h.

$x^4+x^3+x+1=(x^4+x^3)+(x+1)$
$=x^3(x+1)+(x+1)=(x^3+1)(x+1)=(x+1)(x^2-x+1)(x+1)$

$=(x+1)^2(x^2-x+1)$

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

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

<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)

a: Xét ΔABC có BM/BC=BD/BA

nên MD//AC

=>MM' vuông góc AB

=>M đối xứngM' qua AB

b: Xét tứ giác AMBM' có

D là trung điểm chung của AB và MM'

MA=MB

Do đó: AMBM' là hình thoi