Bài 1: Mở đầu về phương trình

Bài này phải giải phương trình bậc 2 mà ghi là lớp 8 cũng khó!

\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-3\right)-24=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)-14=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)^2-7\left(2x^2+3x-1\right)+2\left(2x^2+3x-1\right)-14=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(2x^2+3x-1-7\right)+2\left(2x^2+3x-1-7\right)=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(2x^2+3x-8\right)+2\left(2x^2+3x-8\right)=0\)

\(\Leftrightarrow\left(2x^2+3x-1+2\right)\left(2x^2+3x-8\right)=0\)

\(\Leftrightarrow\left(2x^2+3x+1\right)\left(2x^2+3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x+1=0\\2x^2+3x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{-3+\sqrt{73}}{4}\\x=\dfrac{-3-\sqrt{73}}{4}\end{matrix}\right.\)

pt có 4 nghiệm.

(x+2)(x+3)(x+8)(x+12)=4x²

Ta nhóm như sau: [(x+2)(x+12)][(x+3)(x+8)]=4x²

<=> (x² + 14x + 24)(x² + 11x +24) = 4x²

Vì x = 0 , không phải nghiệm của pt nên chia cả hai vế của pt cho x² \(\ne\) 0, ta có:

\(\left(x+14+\dfrac{24}{x}\right)\left(x+11+\dfrac{24}{x}\right)=4\) .

Đặt: \(x+\dfrac{24}{x}=y\) , ta có: (y+14)(y+11)-4=4

<=> y² + 24y+150 = 0

Giải pt ta được y₁ = -10 ; y₂ = -15 \(\Rightarrow\left[{}\begin{matrix}x^2+10x+24=0\\x^2+15x+24=0\end{matrix}\right.\)

Pt có 4 nghiệm x₁ = -4 ; x₂ = -6 ; x_3,4 = \(\dfrac{-15\pm\sqrt{129}}{2}\)

\(\left(x^2+5x\right)^2-2x^2-10x=24\)

\(\Leftrightarrow\left[x\left(x+5\right)\right]^2-2x\left(x+5\right)-24=0\)

\(\Leftrightarrow\left[x\left(x+5\right)\right]^2-2x\left(x+5\right)+1-25=0\)

\(\Leftrightarrow\left[x\left(x+5\right)-1\right]^2-5^2=0\)

\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\left(x+1\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\Leftrightarrow x=1\\x+6=0\Leftrightarrow x=-6\\x+1=0\Leftrightarrow x=-1\\x+4=0\Leftrightarrow x=-4\end{matrix}\right.\)

\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10+3x^2+10x-8\right)\left(5x^2-2x+10-3x^2-10x+8\right)=0\)

\(\Leftrightarrow\left(8x^2+8x+2\right)\left(2x^2-12x+18\right)=0\)

\(\Leftrightarrow2\left(4x^2+4x+1\right).2\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow4\left(2x+1\right)^2\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy ..........

\(-16-k^2+2k=0\Leftrightarrow-k^2+2k-16=0\Leftrightarrow-\left(k^2-2k+16\right)=0\)

\(\Leftrightarrow k^2-2k+16=0\Leftrightarrow k^2-2k+1+15=0\Leftrightarrow\left(k-1\right)^2+15=0\)

ta có : \(\left(k-1\right)^2\ge0\) với mọi giá trị của \(k\) \(\Rightarrow\left(k-1\right)^2+15\ge15>0\)

vậy phương trình vô nghiệm

\(\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]=9x^2\)

\(\Leftrightarrow\left(x^2-12x+20\right)\left(x^2-9x+20\right)=9x^2\) (*)

Đặt \(t=x^2-\dfrac{21}{2}x+20\)

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

\(\Leftrightarrow t^2-\dfrac{9}{4}x^2=9x^2\)

\(\Leftrightarrow t^2=\dfrac{45}{4}x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3\sqrt{5}}{2}x\\t=-\dfrac{3\sqrt{5}}{2}x\end{matrix}\right.\)

Đến đây Mai tự làm tiếp nha ^^

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)

Ta có: \(A=x^2\left(x^4-1\right)=x^2.\left[\left(x^2\right)^2-1\right]=x^2\left(x^2-1\right)\left(x^2+1\right)\)

\(=\left(x^2-1\right)x^2\left(x^2+1\right)⋮6\) (vì tích 3 số tự nhiên liên tiếp luôn chia hết cho 6) \(\Rightarrow A⋮6\) => đpcm

b) \(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

a: Sửa đề: \(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)

\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1-4\right)-16=0\)

\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1\right)-4=0\)

\(\Leftrightarrow\left(2x^2-3x-1-4\right)\left(2x^2-3x-1+1\right)=0\)

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

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

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)x\left(2x-3\right)=0\)

hay \(x\in\left\{\dfrac{5}{2};-1;0;\dfrac{3}{2}\right\}\)

b: \(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

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

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

hay \(x\in\left\{-2;1\right\}\)