Câu 2:

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

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

\(=3\)

Câu 3: a)Có  \(x=y+5\Rightarrow x-y=5\)

 \(x\left(x+2\right)+y\left(y-2\right)-2xy+65\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+65\)\(=\left(x-y\right)^2+2\left(x-y\right)+65\)

\(=5^2+2.5+65=100\)

b)\(x^2+y\left(y-2x\right)+75=x^2-2xy+y^2+75\)

\(=\left(x-y\right)^2+75=5^2+75\)

\(=100\)

Bài 4:

a) \(A=x^3-30x^2-31x+1\)

=\(x^3-\left(31-1\right)x^2-31x+1\)\(=x^3-\left(x-1\right)x^2-x^2+1\)\(=x^3-x^3+x^2-x^2+1=1\)

b)\(B=x^5-15x^4+16x^3-29x^2+13x\)

\(=x^5-\left(14+1\right)x^4+\left(14+2\right)x^3-\left(2.14+1\right)x^2+\left(14-1\right)x\)

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

\(=x^5-x^5+x^4+x^4+2x^3-2x^3-x^2+x^2-x\)

\(=-x=-14\)

Bài 5

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

\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)

\(=5n^2+5n=5n\left(n+1\right)⋮5\)

\(b.\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-\left(6n^2-3n+10n-5\right)\)

\(=24n+10\) \(=2\left(12n+5\right)⋮2\)

\(A=\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

\(=\left(\sqrt{5}-1-2\right).\left[\dfrac{4\left(\sqrt{5}-1\right)}{\left(1+\sqrt{5}\right)\left(\sqrt{5}-1\right)}+4\right]\)

\(=\left(\sqrt{5}-3\right)\left[\dfrac{4\left(\sqrt{5}-1\right)}{5-1}+4\right]\)\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)=-4\)

B=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\) ( \(a\ge0,a\ne1\))

\(=\dfrac{a+1-2\sqrt{a}}{a+1}:\left[\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+\left(a+1\right)}\right]\)

\(=\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}:\left[\dfrac{a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\right]\)

\(=\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}:\dfrac{a-2\sqrt{a}+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\)\(=\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}.\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2}\)

\(=\sqrt{a}+1\)

b, \(B=\sqrt{a}+1=\sqrt{2022-2\sqrt{2021}}+1\)\(=\sqrt{2021-2\sqrt{2021}+1}+1\)

\(=\sqrt{\left(\sqrt{2021}-1\right)^2}+1=\)\(\left|\sqrt{2021}-1\right|+1\)\(=\sqrt{2021}-1+1=\sqrt{2021}\)

Bài 4: 

Gọi \(x_0\)  là nghiệm chung của hai pt

Ta có hệ: \(\left\{{}\begin{matrix}x_0^2-\left(m+4\right)x_0+m+5=0\\x_0^2-\left(m+2\right)x_0+m+1=0\end{matrix}\right.\)

Trừ vế cho vế \(\Rightarrow-2x_0+4=0\)

\(\Leftrightarrow x_0=2\)

Thay x₀=2 vào pt (1) ta được: \(4-2\left(m+4\right)+m+5=0\) 

\(\Leftrightarrow m=1\)

Vậy m=1

Bài 5:

Gọi \(x_0\) là nghiệm chung của hai pt

Có hệ:\(\left\{{}\begin{matrix}2x_0^2+mx_0-1=0\\mx_0^2-x_0+2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}mx_0=1-2x_0^2\left(1\right)\\mx_0^2-x_0+2=0\left(2\right)\end{matrix}\right.\)

Thay (1) vào (2) ta được: \(\left(1-2x_0^2\right)x_0-x_0+2=0\)

\(\Leftrightarrow-2x_0^3+2=0\) \(\Leftrightarrow x_0=1\)

Thay \(x_0=1\) vào (1) ta tìm được m=-1

Vậy m=-1

Bài 1:

a)\(x^2+2\left(1-m\right)x-m=0\)

Xét \(\Delta=4\left(1-m\right)^2-4\left(-m\right)=4m^2-4m+4\)\(=\left(2m-1\right)^2+3>0\forall m\)

=>Pt luôn có hai nghiệm pb với mọi m

b)\(x^2+mx-m^2-1=0\)

Xét \(\Delta=m^2-4\left(-m^2-1\right)=5m^2+4>0\forall m\)

=>Pt luôn có hai nghiệm pb với mọi m

Bài 2:\(mx^2-\left(2m-1\right)x+2=0\)

Để pt có nghiệm kép \(\Leftrightarrow\Delta=0\)

\(\Leftrightarrow\left(2m-1\right)^2-4m.2=0\)

\(\Leftrightarrow4m^2-12m+1=0\)

Có \(\Delta_m=128>0\)

\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{3+2\sqrt{2}}{2}\\m=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\)

Bài 3:

Gọi \(x_0\) là nghiệm chung của hai pt, ta có hệ:

\(\left\{{}\begin{matrix}x_0^2+mx_0+2=0\\x_0^2+2x_0+m=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_0^2+mx_0+2=0\left(1\right)\\m=-x^2_0-2x_0\left(2\right)\end{matrix}\right.\)

Thay (2) vào (1) ta được: \(x_0^2+\left(-x^2_0-2x_0\right)x_0+2=0\)

\(\Leftrightarrow x_0^3+x^2_0-2=0\)\(\Leftrightarrow\left(x_0-1\right)\left(x_0^2+2x_0+2\right)=0\)

\(\Leftrightarrow x_0=1\) thay x₀=1 vào pt (1) ta tìm được m=-3

Vậy m=-3

\(f\left(x\right)=x^2-\left(2m-1\right)x-2\sqrt{x}+m^2-2m\)

1) đk: \(x\ge0\)

Thay m=2 vào \(f\left(x\right)=0\) ta được: \(x^2-3x-2\sqrt{x}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\\\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)(tm)

2) Thay m=6 vào f(x) ta được:

\(f\left(x\right)=x^2-11x-2\sqrt{x}+24\)

\(f\left(x\right)\le0\) \(\Leftrightarrow x^2-11x-2\sqrt{x}+24\le0\) (bước này coi \(\sqrt{x}\) là nghiệm thì sẽ trở thành bpt

bậc 4 ,bấm máy tính sẽ tìm được nghiệm)

\(\Leftrightarrow\)\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-\dfrac{-1+\sqrt{17}}{2}\right)\left(\sqrt{x}+\dfrac{1+\sqrt{17}}{2}\right)\le0\) 

mà \(\sqrt{x}\ge0\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+2>0\\\sqrt{x}+\dfrac{1+\sqrt{17}}{2}>0\end{matrix}\right.\)

bpt \(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}-\dfrac{-1+\sqrt{17}}{2}\right)\le0\)

\(\Leftrightarrow\sqrt{x}\in\left[\dfrac{-1+\sqrt{17}}{2};3\right]\)

\(\Leftrightarrow x\in\left[\dfrac{9-\sqrt{17}}{2};9\right]\)

Áp dụng bđt bunhia có:

\(\left(x^2+4y^2\right)\left(1+\dfrac{1}{4}\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow\dfrac{25}{4}\ge\left(x+y\right)^2\)\(\Leftrightarrow x+y\le\dfrac{5}{2}\)

Dấu = xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\x^2+4y^2=5\end{matrix}\right.\Leftrightarrow\) \(\left\{{}\begin{matrix}16y^2+4y^2=5\\x=4y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Để pt có hai nghiệm pb:

\(\Leftrightarrow\)\(\Delta=16-4\left(m-4\right)>0\)\(\Leftrightarrow8>m\)

Có\(\left(x_1-1\right)\left(x_2^2-3x_2+m-3\right)=-2\)

\(\Leftrightarrow\left(x_1-1\right)\left(x^2_2-4x_2+m-4\right)+\left(x_1-1\right)\left(x_2+1\right)=-2\)

\(\Leftrightarrow x_1x_2+x_1-x_2-1=-2\) (*) (vì x₂ là một nghiệm của pt nên \(x_2^2-4x_2+m-4=0\))

TH1: \(x_1>x_2\)

(*)\(\Leftrightarrow x_1x_2+\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}+1=0\)

\(\Leftrightarrow m-4+\sqrt{4^2-4\left(m-4\right)}+1=0\)

\(\Leftrightarrow\sqrt{32-4m}=3-m\) \(\Leftrightarrow\left\{{}\begin{matrix}32-4m=9-6m+m^2\\m\le3\end{matrix}\right.\) \(\Leftrightarrow m=1-2\sqrt{6}\)

TH2:\(x_1< x_2\)

(*)\(\Leftrightarrow\)\(x_1x_2-\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}+1=0\)

\(\Leftrightarrow m-4+1=\sqrt{32-4m}\) \(\Leftrightarrow\left\{{}\begin{matrix}m-3\ge0\\\left(m-3\right)^2=32-4m\end{matrix}\right.\)\(\Leftrightarrow m=1+2\sqrt{6}\) (tm đk m<8)

Vậy \(\left[{}\begin{matrix}m=1-2\sqrt{6}\\m=1+2\sqrt{6}\end{matrix}\right.\)

14)\(\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin\dfrac{\pi}{4}-sin2x}{sin\dfrac{\pi}{4}+sin2x}\)\(=\dfrac{2.sin\left(\dfrac{\pi}{4}-x\right)cos\left(\dfrac{\pi}{4}+x\right)}{2sin\left(\dfrac{\pi}{4}+x\right)cos\left(\dfrac{\pi}{4}-x\right)}\)\(=tan\left(\dfrac{\pi}{4}-x\right)cot\left(\dfrac{\pi}{4}+x\right)\)\(=tan\left[\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+x\right)\right]cot\left(\dfrac{\pi}{4}+x\right)\)

\(=cot\left(\dfrac{\pi}{4}+x\right).cot\left(\dfrac{\pi}{4}+x\right)\)

\(=cot^2\left(\dfrac{\pi}{4}+x\right)\)

15)\(\dfrac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}\)\(=\dfrac{\left(sinx+sin5x\right)+sin3x}{\left(cosx+cos5x\right)+cos3x}\)\(=\dfrac{2sin3x.cos\left(-2x\right)+sin3x}{2cos3x.cos\left(-2x\right)+cos3x}=\dfrac{sin3x\left[2cos\left(-2x\right)+1\right]}{cos3x\left[2cos\left(-2x\right)+1\right]}\)\(=tan3x\)

16)\(\dfrac{cos5x-cosx}{sin4x+sin2x}=\dfrac{-2.sin3x.sin2x}{2.sin3x.cosx}\)\(=\dfrac{-2.sinx.cosx}{cosx}\)\(=-2.sinx\)

17)\(\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)}\)\(=\dfrac{cos^2x-\left(sin^2x+cos^2x\right)\left(cos^2x-sin^2x\right)}{2\left(1-cosx\right)}\)\(=\dfrac{cos^2x-\left(cos^2x-sin^2x\right)}{2\left(1-cosx\right)}\)\(=\dfrac{sin^2x}{2\left(1-cosx\right)}\)

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

18) (Xem lại đề)

19)\(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)\(=\dfrac{\left(1+cos2x\right)+\left(cosx+cos3x\right)}{2cos^2x+cosx-1}\)\(=\dfrac{2cos^2x+2.cos2x.cosx}{2cos^2x+cosx-1}\)

\(=\dfrac{2cosx\left(cosx+cos2x\right)}{2cos^2x+cosx-1}\)\(=\dfrac{2cosx\left(cosx+2cos^2-1\right)}{2cos^2x+cosx-1}\)\(=2cosx\)

b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)

\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)

\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)

\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)

Đặt \(t=\sqrt{2+x}+\sqrt{4-x}\)  (\(t\in\left[\sqrt{6};2\sqrt{3}\right]\) )      

\(\Leftrightarrow t^2=6+2\sqrt{8+2x-x^2}\)

\(\Leftrightarrow\dfrac{t^2-6}{2}=\sqrt{8+2x-x^2}\)

Khi đó ta cần tìm m để bpt \(t-\dfrac{t^2-6}{2}\le m\) có nghiệm \(t\in\left[\sqrt{6};2\sqrt{3}\right]\)

\(\Leftrightarrow-t^2+2t+6-2m\le0\) có nghiệm  \(t\in\left[\sqrt{6};2\sqrt{3}\right]\)

Đặt \(f\left(t\right)=-t^2+2t+6-2m\) , \(t\in\left[\sqrt{6};2\sqrt{3}\right]\)

BBT 

TH1: \(maxf\left(t\right)\le0\) \(\Leftrightarrow f\left(1\right)\le0\) \(\Leftrightarrow7-2m\le0\) \(\Leftrightarrow m\ge\dfrac{7}{2}\)       (I)

TH2: \(maxf\left(t\right)>0\Leftrightarrow7-2m>0\Leftrightarrow m< \dfrac{7}{2}\)

Để \(f\left(t\right)\le0\) có nghiệm \(t\in\left[\sqrt{6};2\sqrt{3}\right]\)

 \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{6}-2m\le0\\2\sqrt{6}-2m>0\ge-6+4\sqrt{3}-2m\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\ge\sqrt{6}\\\sqrt{6}>m\ge-3+2\sqrt{3}\end{matrix}\right.\)

Kết hợp với đk ta có:\(\left[{}\begin{matrix}\dfrac{7}{2}>m\ge\sqrt{6}\\\sqrt{6}>m\ge-3+2\sqrt{3}\end{matrix}\right.\)           (II)

Từ (I) (II) ta có: \(m\in\left[-3+2\sqrt{3};+\infty\right]\)