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

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

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

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=\(\dfrac{-23}{7}\)

\(a)\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\\ \Leftrightarrow15x-63x^2-15+63x+63x^2-35x+36x-20=44\\ \Leftrightarrow79x-35=44\\ \Leftrightarrow79x=79\Rightarrow x=1\)

\(b)\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\\ \Leftrightarrow x^2+2x+x+2\left(x+5\right)-x^3-8x^2=27\\ \Leftrightarrow x^2\left(x+5\right)+2x\left(x+5\right)+x\left(x+5\right)+2\left(x+5\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+5x^2+2x^2+10x+x^2+5x+2x+10-x^3-8x^2=27\\ \Leftrightarrow17x+10=27\\ \Leftrightarrow17x=17\Rightarrow x=1\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

\(3x-2x^2=0\)

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

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