a) = 8x-6x-20+15x-15-6x+55-22x-30x+75

b)3x³+5x²-6x²+10x+9x-15-2x³+7x²+2x²+7x-2x-7

bạn chỉ cần chép lại đầu bài rồi ghi kết quả này vào là xong nhá bởi vì đây là nhân đa thức với đa thức mà

a) (2x-5)(4-3x)-(3x+11)(5-2x)-15(2x-5)

  =(2x-5)(4-3x)+(3x+11)(2x-5)-15(2x-5)

  =(2x-5)[(4-3x)+(3x+11)-15]

  =(2x-5)(4-3x+3x+11-15)

  = 2x-5

\(6\cdot x+9=15\)

\(\Rightarrow6\cdot x=15-9\)

\(\Rightarrow6\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{6}=1\)

_______________

\(43+2x=49\)

\(\Rightarrow2x=49-43\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=\dfrac{6}{2}=3\)

____________

\(x:2+5=11\)

\(\Rightarrow x:2=11-5\)

\(\Rightarrow x:2=6\)

\(\Rightarrow x=6\cdot2=12\)

______________

\(77-11x=0\)

\(\Rightarrow11x=77\)

\(\Rightarrow x=\dfrac{77}{11}\)

\(\Rightarrow x=7\)

_______________

\(12-4:x=8\)

\(\Rightarrow4:x=12-8\)

\(\Rightarrow4:x=4\)

\(\Rightarrow x=\dfrac{4}{4}=1\)

_____________

\(x:3+8=11\)

\(\Rightarrow x:3=11-8\)

\(\Rightarrow x:3=3\)

\(\Rightarrow x=3\cdot3=9\)

a: 6x+9=15

=>6x=6

=>x=1

b: 2x+43=49

=>2x=6

=>x=3

c: x:2+5=11

=>x:2=6

=>x=12

d: 77-11x=0

=>7-x=0

=>x=7

e: 12-4:x=8

=>4:x=4

=>x=1

f: x:3+8=11

=>x:3=3

=>x=9

a.x=1

b.x=3

c.x=12

d.x=7

e.x=1

f.x=9

1) \(2x\cdot\left(x-3\right)-5=3x\left(2x-5\right)-4x^2+40\)

\(\Leftrightarrow2x^2-6x-5=6x^2-15x-4x^2+40\)

\(\Leftrightarrow2x^2-6x-5=2x^2-15x+40\)

\(\Leftrightarrow2x^2-6x-5-2x^2+15x-40=0\)

\(\Leftrightarrow9x-45=0\)

<=> x=5

2) x(2x-1)-5(-7)²=2x²-2x+5

<=> 2x²-x-5.49=2x²-2x+5

<=> 2x²-x-245-2x²+2x-5=0

<=> x-250=0

<=> x=250

3) |a-2|=10

\(\Leftrightarrow\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-8\end{cases}}}\)

4) |x|=-5

=> Không tồn tại giá trị của x thỏa mãn vì |x| >=0 với mọi x thuộc Z

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

a) Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left(2x+7\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)

\(\Leftrightarrow\left(x+4\right)\cdot\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-4;-\dfrac{10}{3}\right\}\)

b) Ta có: \(\left(4x+14\right)^2=\left(7x+2\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(7x+2\right)^2=0\)

\(\Leftrightarrow\left(4x+14-7x-2\right)\left(4x+14+7x+2\right)=0\)

\(\Leftrightarrow\left(-3x+12\right)\left(11x+16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+12=0\\11x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-12\\11x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{16}{11}\end{matrix}\right.\)Vậy: \(S=\left\{4;-\dfrac{16}{11}\right\}\)

(2x+7)²=(x+3)²

=>(2x+7)²-(x+3)²=0

=>(2x+7-x-3)(2x+7+x+3)=0

=>(x-4)(3x+10)=0

=>x-4=0 hoặc 3x+10=0

TH1:x-4=0=>x=4

TH2:3x+10=0=>x=-10/3

(4x+14)²=(7x+2)²

(4x+14)²-(7x+2)²=0

(4x+14-7x-2)(4x+14+7x+2)=0

(-3x+12)(11x+16)=0

TH1:-3x+12=0=>x=4

TH2:11x+16=0=>x=-16/11

Đề bài là gì bạn nhỉ ?

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

\(=x\left(x^2-16\right)\)

\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^3-16x-x^4+1\)

b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=-7x^2+7x\)

c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-8x^2+20x-8\)

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

a)  x(x+4)(x-4)-(x²+1)(x²-1)

=>x(x²-4²)-(x⁴-1²)

=>x³-16x-x⁴+1

=>-x⁴-x³-15x

b)  7x(4y-x)+4y(y-7x)-2(2y²-3.5x)

=>28xy-7x²+4y²-28xy-4y²+30x

=>-7x²+30x

c)  (3x+1)(2x-5)-4(2x²-5x+2)

=>6x²-15x+2x-5-8x²+20x-8

=>-2x²+7x-13

a) (x - 5)(x - 3) + 2(x - 5) = 0

(x - 5)(x - 3 + 2) = 0

(x - 5)(x - 1) = 0

x - 5 = 0 hoặc x - 1 = 0

*) x - 5 = 0

x = 5

*) x - 1 = 0

x = 1

Vậy x = 1; x = 5

b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)

x³ - 8 - x³ - 8 = 2x + 4

2x = -8 - 8 - 4

2x = -20

x = -20 : 2

x = -10

a)

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

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

\(\left(x-5\right)\left(x-1\right)=0\)

\(x-5=0\) hoặc \(x-1=0\)

+) \(x-5=0\\ \Rightarrow x=5\)

+) \(x-1=0\\ \Rightarrow x=1\)

Vậy \(x=1\) hoặc \(x=5\)

b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)

\(x^3-8-x^3-8=2x+4\)

\(2x=-8-8-4\)

\(2x=-20\)

 \(x=-20:2\)

 \(x=-10\)

Vậy \(x=-10\)

a, \(2x-3< 0\Leftrightarrow2x< 3\Leftrightarrow x< \frac{3}{2}\)

b, \(\left(2x-4\right)\left(9-3x\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}2x-4>0\\9-3x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x< 3\end{cases}\Leftrightarrow2< x< 3}}\)

a. \(2x-3< 0\Leftrightarrow2x< 3\Leftrightarrow x< \frac{3}{2}\)

b. \(\left(2x-4\right)\left(9-3x\right)>0\Leftrightarrow18x-6x-36+12x>0\Leftrightarrow24x>36\Leftrightarrow x>\frac{3}{2}\)

c. \(\frac{2}{3}x-\frac{3}{4}>0\Leftrightarrow\frac{2}{3}x>\frac{3}{4}\Leftrightarrow x>\frac{9}{8}\)

d. \(\left(\frac{3}{4}-2x\right)\left(\frac{-3}{5}+\frac{2}{-61}-\frac{17}{51}\right)\le0\)

\(\Leftrightarrow\frac{3}{4}-2x\le0\Leftrightarrow2x\le\frac{3}{4}\Leftrightarrow x\le\frac{3}{8}\)

e. \(\left(\frac{3}{2}x-4\right).\frac{5}{3}>\frac{15}{6}\Leftrightarrow\frac{3}{2}x-4>\frac{3}{2}\Leftrightarrow\frac{3}{2}x>\frac{11}{2}\Leftrightarrow x>\frac{11}{3}\)

\(a.2x+7x+x=-270\)

\(10x=-270\)

\(x=-27\)

\(b,\left(x-1\right)\left(x-9\right)=0\)

\(=>x-1=0\) \(=>x=1\)

       \(x-9=0=>x=9\)

Vậy \(x\in\left\{1;9\right\}\)