Đăng ít thôi.

Bài 2:

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)

\(=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}\)

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

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

\(=\dfrac{x-3}{x}\)

b) \(\dfrac{6x-3}{x}:\dfrac{4x^2-1}{3x^2}\)

\(=\dfrac{6x-3}{x}.\dfrac{3x^2}{4x^2-1}\)

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

\(=\dfrac{9x}{2x+1}\)

c) \(\dfrac{x+2}{3x}+\dfrac{x-5}{5x}-\dfrac{x+8}{4x}\)

\(=\dfrac{20x\left(x+2\right)+12x\left(x-5\right)-15x\left(x+8\right)}{60x}\)

\(=\dfrac{20x^2+40x+12x^2-60x-15x^2-120x}{60x}\)

\(=\dfrac{17x^2-140x}{60x}\)

d) \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}\)

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

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

\(=\dfrac{3}{x}\).

Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

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

nhiều quá bạn ạ

hay bạn tìm hiểu cách thức chung làm dạng bài tìm GTNN chứ như thế này thì làm lâu lắm

a: \(=49x^2-28x+4+21=\left(7x-2\right)^2+21>=21\)

Dấu '=' xảy ra khi x=2/7

b: \(=8\left(x^2-\dfrac{7}{2}x-\dfrac{1}{8}\right)\)

\(=8\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{51}{16}\right)\)

\(=8\left(x-\dfrac{7}{4}\right)^2-\dfrac{51}{2}>=-\dfrac{51}{2}\)

Dấu '=' xảy ra khi x=7/4

c: \(C=\left(2x^2+5\right)^2+10>=25+10=35\)

Dấu '=' xảy ra khi x=0

2)

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

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy x=0 ; x=-1 ; x=1

b) \(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

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

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

1)

a) \(\left(x-2\right)\left(x^2+3x+4\right)\)

\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)

\(\Leftrightarrow x^3+x^2-2x-8\)

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

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

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

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

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

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

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

\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)

\(=17x^2+5x-6-6x^3\)

pạn ơi pạn đã lm đk chưa? nếu lm đk oy cho mk xem cách lm bài 2 nhé. cảm ơn pạn nhìu lắm

1) \(\left(x-3\right)\left(x-5\right)+44\)

\(=x^2-3x-5x+15+44\)

\(=x^2-8x+59\)

\(=x^2-2.x.4+4^2+43\)

\(=\left(x-4\right)^2+43\ge43>0\)

\(\rightarrowĐPCM.\)

2) \(x^2+y^2-8x+4y+31\)

\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)

\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)

\(\rightarrowĐPCM.\)

3)\(16x^2+6x+25\)

\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)

\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)

\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)

\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)

-> ĐPCM.

4) Tương tự câu 3)

5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)

\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)

\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)

-> ĐPCM.

6) Tương tự câu 5)

7) 8) 9) Tương tự câu 3).

Bài 1a/

\(\frac{1}{1+x+xy}=\frac{xyz}{xyz+x+xy}=\frac{yz}{1+y+yz}\)

\(\frac{1}{1+z+xz}=\frac{y}{y+yz+xyz}=\frac{y}{1+y+yz}\)

Vậy \(M=\frac{1}{1+y+yz}+\frac{y}{1+y+yz}+\frac{yz}{1+y+yz}=1\)

Chiều về làm tiếp

Bài 1b:Lời giải này chủ yếu nhờ dự đoán trước Min là 2011/2012 đạt được khi x=2012

Ta có \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\ge\frac{2011x^2}{2012x^2}=\frac{2011}{2012}\)

Bài 2: Dùng phân tích thành bình phương

\(10x^2+y^2+4z^2+6x-4y-4xz+5=\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)\)

\(=\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}}\)

Bài 3:

a/\(pt\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\Leftrightarrow x=-6,x=5\)

b/ta phân tích vế trái thành:\(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)