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

\(=-3x.\left(x^2+2.x.2+2^2\right)+\left(x^2+x+3x-3\right).\left(x+1\right)-\left(2x\right)^2-2.2.x.\left(-3\right)+\left(-3\right)^2\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+\left(x+3x\right)-3\right).\left(x+1\right)-4x+12x+9\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+4x-3\right)\left(x+1\right)-4x+12x+9\)

\(=-3x^3-12x^2-12x+x^3+4x^2-3x+x^2+4x-3-4x+12x+9\)

\(=\left(-3x^3-x^3\right)+\left(-12x^2+4x^2+x^2\right)+\left(-12x-3x+4x-4x+12x\right)+\left(-3+9\right)\)

\(=-2x^3-7x^2-3x+6\)

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

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

\(=\left(x.x+x.3-3.x+\left(-3\right).3\right)\left(x+2\right)-\left(x.x^2+x.\left(-3\right)-1.x^2+\left(-1\right).\left(-3\right)\right)-5x.x+\left(-5x\right).4-x^2-2x5+5^2\)

\(=\left(x^2+3x-3x-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2+\left(3x-3x\right)-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=x^3+2x^2-9x-15-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^3-x^3\right)+\left(2x^2-x^2-5x^2-x^2\right)+\left(-9x-3x-20x-10x\right)+\left(-18+3+25\right)\)

\(=-5x^2-42x+10\)

HOC24 có câu rất hay :Người hay giúp bạn khác trả lời bài tập sẽ trở thành học sinh giỏi. Người hay hỏi bài thì không. Còn bạn thì sao? đúng tính bà đó . Lên lớp đừng đập nha :)

a) 3 . ( 1/2 - x ) + 1/3 = 7/6 - x

=> 3/2 - 3x + 1/3 = 7/6-x

=> -3x +x=7/6 - 3/2 - 1/3

=> -2x = -2/3

=> x=-2/3 : (-2) = 1/3

hết :)

Bài 2: 

a: \(P=\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=\dfrac{1}{x+2}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=\dfrac{1}{x+2}:\left(\dfrac{4}{x-2}-\dfrac{1}{x+2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\dfrac{1}{x+2}:\dfrac{4x+6-x+2-x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x+8}=\dfrac{x-2}{2x+8}\)

b: Để P=0 thì x-2=0

hay x=2(loại)

Để P=1 thì 2x+8=x-2

hay x=-10(nhận)

Để P>0 thì \(\dfrac{x-2}{2x+8}>0\)

=>x>2 hoặc x<-4

a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{-x+9}\)

\(=\dfrac{5\left(\sqrt{x}-3\right)}{x-9}=\dfrac{5}{\sqrt{x}+3}\)

b: Để A<1 thì A-1<0
\(\Leftrightarrow5-\sqrt{x}-3< 0\)

=>2-căn x<0

=>căn x>2

=>x>4

Q = \(\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)để Q đạt GTLN => \(x^2+1phảiNhỏnhất\)

\(x^2+1\ge1=>x^2+1\)đạt GTNN là 1 khi x=0

vậy Q đạt GTLN =3 khi x = 0

\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^3+x^2\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\\ Q=\dfrac{3\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)}\\ Q=\dfrac{3}{x^2+1}\\ Do\text{ }x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\\ Q=\dfrac{3}{x^2+1}\le3\forall x\\ \text{Dấu “=” xảy ra khi : }\\ x^2=0\\ \Leftrightarrow x=0\\Vậy\text{ }Q_{\left(Max\right)}=3\text{ }khi\text{ }x=0\)

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

\(=x^2-6x+9-2\left(4-x^2\right)-x^2-2x-1\)

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

Viết phép tính k hiểu

a: \(\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\)

\(=\left(\dfrac{x}{x+3}-1\right):\dfrac{9-x^2+x^2-9-x^2+4}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x+3\right)\left(x-2\right)}{-x^2+4}\)

\(=\dfrac{-3}{1}\cdot\dfrac{\left(x-2\right)}{-\left(x-2\right)\left(x+2\right)}=\dfrac{3}{x+2}\)

b: \(\left(\dfrac{x+2}{x-1}-\dfrac{x-1}{x+2}+\dfrac{x+5}{x^2+x-2}\right)\cdot\left(x^3-1\right)\)

\(=\dfrac{x^2+4x+4-x^2+2x-1+x+5}{\left(x+2\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{1}\)

\(=\dfrac{7x+8}{\left(x+2\right)}\cdot\dfrac{x^2+x+1}{1}\)