a: \(3x\left(4x^2-x+7\right)\)

\(=3x\cdot4x^2-3x\cdot x+3x\cdot7\)

\(=12x^3-3x^2+21x\)

b: (2x-1)(4x+5)

\(=8x^2+10x-4x-5\)

\(=8x^2+6x-5\)

c: Sửa đề; \(\left(12x^7-9x^5+6x^3\right):3x^2\)

\(=\frac{12x^7}{3x^2}-\frac{9x^5}{3x^2}+\frac{6x^3}{3x^2}\)

\(=4x^5-3x^3+2x\)

7/3x - 4/9x = 5/6x - 1/3x + 7/3

7/3x - 4/9x - 5/6x + 1/3x = 7/3

\(\frac{42}{18}x\) - \(\frac{8}{18}x\) - \(\frac{15}{18}x\) + \(\frac{6}{18}x\) = 7/3

34/18x - 15/18x + 6/18x = 7/3

19/18x + 6/18x = 7/3

25/18x = 7/3

x = 7/3 : 25/18

x = 7/3 x 18/25

x = 42/25

Vậy x = 42/25

\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)

\(\Leftrightarrow\frac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x+7\right)}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)

\(\Rightarrow\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)=1-6x^2+9x-9\)

\(\Leftrightarrow4x^2+20x+21-4x^2-12x-5=1-6x^2+9x-9\)

\(\Leftrightarrow8x-16=1-6x^2+9x-9\)

\(\Leftrightarrow8x-16-1+6x^2-9x+9=0\)

\(\Leftrightarrow6x^2-x-8=0\)

Tự làm nốt nha 

Trl

-Bạn chuyên toán thcs làm đúng r nhé !~

Học tốt 

nhé bạn ~

`5/39xx(7 4/5xx1 2/3+8 1/3xx7 4/5)`

`=5/39xx(39/5xx5/3+25/3xx39/5)`

`=5/39xx39/5(5/3+25/3)`

`=1xx30/3`

`=10`

10

ghi rõ các bược hộ em ạ 

a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x

=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x

=-3

vậy...

1: Đặt \(a=9x^2-6x\)

=>\(45x^2-30x=5\left(9x^2-6x\right)=5a\)

\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)

=>\(\sqrt{a+2}+\sqrt{5a+9}=\sqrt{-a+8}\)

=>\(\sqrt{a+2}-1+\sqrt{5a+9}-2=\sqrt{-a+8}-3\)

=>\(\frac{a+2-1}{\sqrt{a+2}+1}+\frac{5a+9-4}{\sqrt{5a+9}+2}=\frac{-a+8-9}{\sqrt{-a+8}+3}\)

=>\(\left(a+1\right)\left(\frac{1}{\sqrt{a+2}+1}+\frac{5}{\sqrt{5a+9}+2}+\frac{1}{\sqrt{a+8}+3}\right)=0\)

=>a+1=0

=>a=-1

=>\(9x^2-6x=-1\)

=>\(9x^2-6x+1=0\)

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

=>3x-1=0

=>3x=1

=>x=1/3

2: Đặt \(x^2-2x=a\)

\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\)

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

=>\(\sqrt{2a+3}+\sqrt{3a+7}=-a+2\)

=>\(\sqrt{2a+3}-1+\sqrt{3a+7}-2=-a+2-3\)

=>\(\frac{2a+2}{\sqrt{2a+3}+1}+\frac{3a+7-4}{\sqrt{3a+7}+2}=-a-1\)

=>\(\left(a+1\right)\left(\frac{2}{\sqrt{2a+3}+1}+\frac{3}{\sqrt{3a+7}+2}+1\right)=0\)

=>a+1=0

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

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

=>x-1=0

=>x=1

Đề bài yêu cầu gì?

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4