a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

Bài 3: 

\(\Leftrightarrow x^3+64-x^3+25x=264\)

hay x=8

\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)

a)\(M=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+23x-55-6x^2-23x-21=-76\)

b) \(N=\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1=9\)

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^4-10x^3+6x^2\)

c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)

d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)

\(a,=x^2-6x+9-x^2+6x=9\\ b,=4x^2+4x+1-4x^2+9-4x-8=2\\ c,=\left(2x^2-2x-x+1\right):\left(x-1\right)\\ =\left(x-1\right)\left(2x-1\right):\left(x-1\right)=2x-1\)

`a)(x-3)^2-x(x-6)`

`=x^2-6x+9-x^2+6x=9`

`b)(2x+1)^2-(3+2x)(2x-3)-4(x+2)`

`=4x^2+4x+1-(4x^2-9)-4x-8`

`=2`

`c)(2x^2-3x+1):(x-1)`

`=(2x^2-2x-x+1):(x-1)`

`=[2x(x-1)-(x-1)]:(x-1)`

`=2x-1`

a) \(\left(x-3\right)^2-x\left(x-6\right)=x^2-6x+9-x^2+6x=9\)

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

c) \(\left(2x^2-3x+1\right):\left(x-1\right)=\left[2x\left(x-1\right)-\left(x-1\right)\right]:\left(x-1\right)=\left[\left(x-1\right)\left(2x-1\right)\right]:\left(x-1\right)=2x-1\)

a) Đa thức thương  x 2  – 6x + 9.

b) Đa thức thương 2 x 2  – 5.

c) Đa thức thương  x 2  + 4x + 3 và đa thức dư -12.

d) Đa thức x + 5 và đa thức dư x – 4.

a, \(\Leftrightarrow2x^2=72\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm6\)

Vậy ...

\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)

\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)

Vậy ...

\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)

Vậy ...

\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)

\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)

Vậy ...

a) 2x² - 72 = 0

\(\Rightarrow\) 2x² = 72

\(\Rightarrow\) x² = 36 = 6² = (- 6)²

\(\Rightarrow\) x = 6 hoặc x = - 6

Vậy x = 6 hoặc x = - 6

b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)

\(\Rightarrow\)  (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)

\(\Rightarrow\)  \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)

\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)

\(\Rightarrow\) x = \(\dfrac{13}{4}\)

Vậy x = \(\dfrac{13}{4}\)

c) \(2x+\dfrac{3}{10}=1\dfrac{5}{6}.\dfrac{6}{11}\)

\(\Rightarrow\) \(2x+\dfrac{3}{10}=\dfrac{11}{6}.\dfrac{6}{11}=1\)

\(\Rightarrow\) \(2x=\dfrac{7}{10}\)

\(\Rightarrow\) \(x=\dfrac{7}{20}\)

Vậy \(x=\dfrac{7}{20}\)

d) \(2\dfrac{1}{4}:\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\Rightarrow\) \(\dfrac{9}{4}:\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}\)

\(\Rightarrow\) \(x-\dfrac{22}{3}=\dfrac{3}{2}\)

\(\Rightarrow\) \(x=\dfrac{53}{6}\)

Vậy \(x=\dfrac{53}{6}\)

Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5