Câu 3a:

4\(x^3\) - 9\(x\)

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

= \(x\) x [(2\(x\))\(^2\) - 3\(^2\)]

= \(x\times\) [2\(x\) - 3][\(2x+3\)]

b; \(x^2+2x-3\)

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

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

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

= (\(x-1)\)(\(x+3\))

Câu c:

\(x^2\) - y\(^2\) - 6\(x\) + 9

= (\(x^2\) - 6\(x\) + 9) - y\(^2\)

= (\(x^2-2.3x\) + 3\(^2\)) - y\(^2\)

= (\(x-3\))\(^2\) - y\(^2\)

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

36 ⋮ \(x\) ; 45 ⋮ \(x\) ; 18 ⋮ \(x\) mà \(x\) lớn nhất nên

\(x\) ∈ ƯCLN(36; 45; 18)

36 = \(2^2.3^2\)

45 = 5.3\(^2\)

18 = 2.3\(^2\)

ƯCLN(36; 45; 18) = 2.3\(^2\) = 2.9 = 18

Vậy \(x=18\)

\(36=2^2\cdot3^2;45=5\cdot3^2;18=2\cdot3^2\)

=>\(ƯCLN\left(36;45;18\right)=3^2=9\)

36⋮x

45⋮x

18⋮x

Do đó: x∈ƯC(36;45;18)

mà x lớn nhất

nên x=ƯCLN(36;45;18)

=>x=9

Sửa đề: \(2a^2+7ab+3b^2=0\)

=>\(2a^2+6ab+ab+3b^2=0\)

=>2a(a+3b)+b(a+3b)=0

=>(a+3b)(2a+b)=0

=>\(\left[\begin{array}{l}a+3b=0\\ 2a+b=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=-3b\\ b=-2a\end{array}\right.\)

TH1: a=-3b

\(\frac{8a-3b}{2a-b}-\frac{2a-5b}{2a+b}\)

\(=\frac{8\cdot\left(-3b\right)-3b}{2\left(-3b\right)-b}-\frac{2\cdot\left(-3b\right)-5b}{2\cdot\left(-3b\right)+b}=\frac{-24b-3b}{-6b-b}-\frac{-6b-5b}{-6b+b}\)

\(=\frac{-27}{-7}-\frac{-11}{-5}=\frac{27}{7}-\frac{11}{5}=\frac{135}{35}-\frac{77}{35}=\frac{58}{35}\)

TH2: b=-2a

\(\frac{8a-3b}{2a-b}-\frac{2a-5b}{2a+b}\)

\(=\frac{8a-3\cdot\left(-2a\right)}{2a-\left(-2a\right)}-\frac{2a-5\cdot\left(-2a\right)}{2a-2a}=\frac{14a}{4a}-\frac{12a}{0a}\)

=>Khi b=-2a thì biểu thức không có giá trị

99 x 100 = 9900

220 : 100 = \(\frac{11}{5}\)

3020 : 1000 = \(\frac{151}{50}\)

690 x 1000 = 690 000

20 : 100 = \(\frac15\)

200 + 12 - (6 x 11)

= 200 + 12 - 66

= 212 - 66

= 146

200 + 12 - (6 x 11)

= 200 + 12 - 66

= 212 - 66

= 146😁😁😁

\(2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+12\cdot4\)

=3+48

=51

\(2^3-5^3:5^2+12\times2^2\)

\(=8-5+12\times4\)

\(=8-5+48=51\)

b; \(\frac56\) x \(x+\) \(\frac25\) = \(\frac13\times x\) + \(\frac{11}{12}\)

\(\frac56\times x\) = \(\frac13\times x\) + \(\frac{11}{12}\) - \(\frac25\)

\(\frac56\times x\) = \(\frac13\times x\) + (\(\frac{55}{60}\) - \(\frac{24}{60}\))

\(\frac56\times x\) = \(\frac13\times x\) + \(\) \(\frac{31}{60}\)

\(\frac56\times x\) - \(\frac13\times x\) = \(\frac{31}{60}\)

\(x\) x(\(\frac56-\frac13\)) = \(\frac{31}{60}\)

\(x\) x (\(\frac56\) - \(\frac26\)) = \(\frac{31}{60}\)

\(x\) x \(\frac12\) = \(\frac{31}{60}\)

\(x\) = \(\frac{31}{60}\) : \(\frac12\)

\(x\) = \(\frac{31}{60}\) x \(\frac21\)

\(x\) = \(\frac{31}{30}\)

\(a,\)

\(\frac45\times x-\frac13=\frac12\times x\)

\(\frac45\times x-\frac12\times x=\frac13\)

\(\left(\frac45-\frac12\right)\times x=\frac13\)

\(\frac{3}{10}\times x=\frac13\)

\(x=\frac13:\frac{3}{10}\)

\(x=\frac{10}{9}\)

Vậy \(x=\frac{10}{9}\)

1: Diện tích mặt sàn là 5x10=50(\(m^2\) )

2: Diện tích mỗi viên gạch là \(50\times50=2500\left(\operatorname{cm}^2\right)=0,25\left(m^2\right)\)

Số viên gạch cần dùng là 50:0,25=200(viên)

Số hộp gạch cần dùng là 200:4=50(hộp)

Số tiền phải trả là:

\(160000\times50=8000000\) (đồng)

3: Số tiền cần có là:

\(8000000\times22=176000000\) (đồng)

\(\left(2x-1\right)\left(4x^2+2x+1\right)-2x\left(2x-3\right)\left(2x+3\right)=81x^2\)

\(8x^3-1-2x\left(4x^2-9\right)=81x^2\)

\(8x^3-1-\left(8x^3-18x\right)=81x^2\)

\(8x^3-1-8x^3+18x=81x^2\)

\(18x-1=81x^2\)

\(81x^2-18x+1=0\)

\(\left(9x\right)^2-2\cdot9x+1=0\)

\(\left(9x-1\right)^2=0\)

\(9x-1=0\)

\(x=\frac19\)

Vậy \(x=\frac19\)

Ta có: \(\left(2x-1\right)\left(4x^2+2x+1\right)-2x\left(2x-3\right)\left(2x+3\right)=81x^2\)

=>\(8x^3-1-2x\left(4x^2-9\right)=81x^2\)

=>\(8x^3-1-8x^3+18x=81x^2\)

=>\(81x^2=18x-1\)

=>\(81x^2-18x+1=0\)

=>\(\left(9x-1\right)^2=0\)

=>9x-1=0

=>9x=1

=>\(x=\frac19\)