Ta có: (3x-3)(5-21x)+(7x+4)(6x-5)=45

\(\Leftrightarrow15x-63x^2-15+63x+42x^2-35x+24x-20-45=0\)

\(\Leftrightarrow-21x^2+67x-80=0\)

\(\Leftrightarrow-21\left(x^2-\frac{67}{21}x-\frac{80}{21}\right)=0\)

mà -21<0

nên \(x^2-\frac{67}{21}x-\frac{80}{21}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{67}{42}+\frac{4489}{1764}-\frac{11209}{1764}=0\)

\(\Leftrightarrow\left(x-\frac{67}{42}\right)^2=\frac{11209}{1764}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{67}{42}=\frac{\sqrt{11209}}{42}\\x-\frac{67}{42}=-\frac{\sqrt{11209}}{42}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{67+\sqrt{11209}}{42}\\x=\frac{67-\sqrt{11209}}{42}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{67+\sqrt{11209}}{42};\frac{67-\sqrt{11209}}{42}\right\}\)

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

F= 21x⁸ - 24x⁶ + 9x⁵ + 3x³ + 6x² + 2006 

  = 3x²( 7x⁶ - 8x⁴ + 3x³ + x +2) +2006 

  = 0 + 2006 

  = 0

sorry cái kquả ban nãy mình viết nhầm

Kquả là 2006

1/

a. \(3x\left(5x^2-2x-1\right)\)

\(=15x^3-6x^2-3x\)

b. \(\left(x^2-2xy+3\right)\left(-xy\right)\)

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

c. \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)

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

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

a) thiếu đề

b) \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)

\(15x-63x^2-15+63x+63x^2-35x+36x-20=44\)

\(79x-35=40\)

\(79x=75\)

\(x=\frac{75}{79}\)

Câu 2 b) nhầm đoạn cuối

sửa từ dòng 3

\(79x-35=44\)

\(79x=79\)

\(x=1\)

a)    x³-x²-21x+45=0

<=> x³+5x²-6x²-30x+9x+45=0

<=> (x+5)(x²-6x+9)=0

<=> (x+5)(x²-3x-3x+9)=0

<=> (x+5)(x-3)²=0

 Vậy S={-5;3}

b)    X³+3X²+4X+2=0

<=>  X³+X²+2X²+2X+2X+2=0

<=> (X+1)(X²+2X+2)=0

VÌ  X²+2X+2 >=0

NÊN S={-1}

C)    X⁴+7X-8=0

<=> X⁴-X³+X³-X²+X²-X+8X-8=0

<=> (X-1)(X³+X²+X+8)=0

VÌ X³+X²+X+8>=0

NÊN S={1}

D)     6X⁴-X³-7X²+X+1=0

<=>  6X⁴-6X³+5X³-5X²-2X²+2X-X+1=0

<=>  (X-1)(6X³+5X²-2X-1)=0

<=> (X-1)(6X³-3X²+8X²-4X+2X-1)=0

<=> (X-1)(2X-1)(3X^2_4X+1)=0

<=>  (X-1)(2X-1)(3X²-3x-x+1)=0

<=> (X-1)²(2X-1)(3x-1)=0

vậy S={1/3;1/2;1}