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)

1 does this red dress cost

2 is this shirt

does this short cost

3 do these shoes cost

4 has brown backpack

5 how to get to the bus stop

6 to stay in bed

7 are cleaned by Nga everyday

8 is played with a bow

9 worked for that company for 10 years

10 15 minutes walking to school

over

camping

was

couldn't

percussion

takes

look after

go around 

go through

wind

how long

for

do

Bài 4:

a: a\(\perp\)c

b\(\perp\)c

Do đó: a//b

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 = \(\dfrac{-34}{9}\)

Đáp án không thôi nhá chứ giải mất tg lắm

a) \(6x^2-8x=2x\left(3x-4\right)\)

b) \(21x^2y^7-12x^5y^3=3x^2y^3\left(7y^4-4x^3\right)\)

c) \(4x^2y-6xy^2=2xy\left(2x-3y\right)\)

d) \(24x^2y-18xy^2=6xy\left(4x-3y\right)\)

e) \(9x^2y^2+12x^2y-15xy^2=3xy\left(3xy+4x-5y\right)\)

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

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

h) \(3x\left(x-y\right)-6y\left(y-x\right)=3x\left(x-y\right)+6y\left(x-y\right)=3\left(x-y\right)\left(x+2y\right)\)

Lời giải:

ĐKĐB $\Rightarrow \frac{2}{c}=\frac{a+b}{ab}\Rightarrow c(a+b)=2ab$

Khi đó:

$\frac{a}{b}-\frac{a-c}{c-b}=\frac{a(c-b)-b(a-c)}{b(c-b)}=\frac{ac-ab-ab+bc}{b(c-b)}=\frac{c(a+b)-2ab}{b(c-b)}=\frac{2ab-2ab}{b(c-b)}=0$

$\Rightarrow \frac{a}{b}=\frac{a-c}{c-b}$ (đpcm)