19.

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

\(\Rightarrow\max\limits_{\left[0;\dfrac{3}{2}\right]}f\left(x\right)=1\)

20.

\(f\left(x\right)< 0;\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

21.

A là đáp án đúng, do đa thức \(f\left(x\right)=-2x^2+3x-4\) có:

\(\left\{{}\begin{matrix}a=-2< 0\\\Delta=3^2-4.\left(-2\right).\left(-4\right)=-23< 0\end{matrix}\right.\)

22.

ĐKXĐ: \(4-x^2\le0\Rightarrow\left(2-x\right)\left(2+x\right)\le0\)

\(\Rightarrow-2\le x\le2\Rightarrow D=\left[-2;2\right]\)

23.

\(f\left(x\right)>0;\forall x\Leftrightarrow\left\{{}\begin{matrix}a=1>0\\\Delta'=\left(2m-3\right)^2-\left(4m-3\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow4m^2-16m+12< 0\)

\(\Rightarrow1< m< 3\)

Bài 3: Phương trình hoành độ giao điểm là:

\(\frac12x^2=2x-m+1\)

=>\(x^2=4x-2m+2\)

=>\(x^2-4x+2m-2=0\)

\(\Delta=\left(-4\right)^2-4\cdot1\cdot\left(2m-2\right)\)

=16-8m+8

=-8m+24

Để (P) cắt (d) tại hai điểm phân biệt thì -8m+24>0

=>-8m>-24

=>m<3

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=4;x_1x_2=\frac{c}{a}=2m-2\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=4^2-2\left(2m-2\right)=16-4m+4=-4m+20\)

\(\left(y_1+y_2\right)\cdot x_1x_2+48=0\)

=>\(\left(\frac12\cdot x_1^2+\frac12\cdot x_2^2\right)\cdot x_1x_2+48=0\)

=>\(\frac12\cdot\left(x_1^2+x_2^2\right)\cdot x_1x_2=-48\)

=>\(x_1x_2\left(x_1^2+x_2^2\right)=-96\)

=>\(\left(-4m+20\right)\left(2m-2\right)=-96\)

=>-4(m-5)*2*(m-1)=-96

=>(m-5)(m-1)=12

=>\(m^2-6m+5-12=0\)

=>\(m^2-6m-7=0\)

=>(m-7)(m+1)=0

=>m=7(loại) hoặc m=-1(nhận)

Bài 1: Phương trình hoành độ giao điểm là:

\(x^2=2x+m-1\)

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

\(\Delta=\left(-2\right)^2-4\cdot1\cdot\left(-m+1\right)=4+4m-4=4m\)

Để (P) cắt (d) tại hai điểm phân biệt thì 4m>0

=>m>0

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2;x_1x_2=\frac{c}{a}=-m+1\)

\(y_1\cdot y_2-x_1x_2=12\)

=>\(\left(x_1x_2\right)^2-x_1x_2=12\)

=>\(\left(-m+1\right)^2-\left(-m+1\right)=12\)

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

=>\(m^2-m-12=0\)

=>(m-4)(m+3)=0

mà m+3>0

nên m-4=0

=>m=4

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2

15.

\(\Delta'=m^2+m-2>0\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -2\end{matrix}\right.\)

Đáp án B

16.

\(\dfrac{\pi}{2}< a< \pi\Rightarrow\dfrac{\pi}{4}< \dfrac{a}{2}< \dfrac{\pi}{2}\Rightarrow\dfrac{\sqrt{2}}{2}< sin\dfrac{a}{2}< 1\Rightarrow\dfrac{1}{2}< sin^2\dfrac{a}{2}< 1\)

\(sina=\dfrac{3}{5}\Leftrightarrow sin^2a=\dfrac{9}{25}\Leftrightarrow4sin^2\dfrac{a}{2}.cos^2\dfrac{a}{2}=\dfrac{9}{25}\)

\(\Leftrightarrow sin^2\dfrac{a}{2}\left(1-sin^2\dfrac{a}{2}\right)=\dfrac{9}{100}\Leftrightarrow sin^4\dfrac{a}{2}-sin^2\dfrac{a}{2}+\dfrac{9}{100}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin^2\dfrac{a}{2}=\dfrac{1}{10}< \dfrac{1}{2}\left(loại\right)\\sin^2\dfrac{a}{2}=\dfrac{9}{10}\end{matrix}\right.\)

\(\Rightarrow sin\dfrac{a}{2}=\dfrac{3\sqrt{10}}{10}\)

17.

Áp dụng công thức trung tuyến:

\(AM=\dfrac{\sqrt{2\left(AB^2+AC^2\right)-BC^2}}{2}=\dfrac{\sqrt{201}}{2}\)

18.

\(\Leftrightarrow x^2+2x+4>m^2+2m\) ; \(\forall x\in\left[-2;1\right]\)

\(\Leftrightarrow m^2+2m< \min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)\)

Xét \(f\left(x\right)=x^2+2x+4\) trên \(\left[-2;1\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-2;1\right]\) ; \(f\left(-2\right)=4\) ; \(f\left(-1\right)=3\) ; \(f\left(1\right)=7\)

\(\Rightarrow\min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)=f\left(1\right)=3\)

\(\Rightarrow m^2+2m< 3\Leftrightarrow m^2+2m-3< 0\)

\(\Rightarrow-3< m< 1\Rightarrow m=\left\{-2;-1;0\right\}\)

Đáp án C

1.

\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

2.

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

3.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)

\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Câu 9: A

Câu 10: C

Câu 11: C

Câu 12: A

Câu 13; B

Câu 14: C

 A

 C

C

 A

 B

C

Câu 8: A

Câu 7: B

Câu 6: B

Câu 5: A

25B

24B

23B

21A
22C