36B

37C

38D

39B

40D

41A

42B

43B

44A

45B

46B

47A

48C

50B

51B

52B

53D

54C

55D

56C

\(\dfrac{12}{16}=\dfrac{132}{176}\\ \dfrac{13}{16}=\dfrac{143}{176}\\ Ta.có:\dfrac{16}{22}< \dfrac{132}{176}< \dfrac{17}{22}< \dfrac{143}{176}< \dfrac{18}{22}\\ Vậy:Chọn.số.17\)

Câu 2: 

Ta có: \(x^2-2\left(m+1\right)x+m^2+4=0\)

a=1; b=-2m-2; \(c=m^2+4\)

\(\text{Δ}=b^2-4ac\)

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

\(=4m^2+8m+4-4m^2-16\)

=8m-12

Để phương trình có hai nghiệm phân biệt thì Δ>0

\(\Leftrightarrow8m>12\)

hay \(m>\dfrac{3}{2}\)

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)=2m+2\\x_1x_2=m^2+4\end{matrix}\right.\)

Vì x1 là nghiệm của phương trình nên ta có: 

\(x_1^2-2\left(m+1\right)\cdot x_1+m^2+4=0\)

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

Ta có: \(x_1^2+2\left(m+1\right)x_2=2m^2+20\)

\(\Leftrightarrow2\left(m+1\right)x_1-m^2-4+2\left(m+1\right)x_2-2m^2-20=0\)

\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-3m^2-24=0\)

\(\Leftrightarrow2\left(m+1\right)\cdot\left(2m+2\right)-3m^2-24=0\)

\(\Leftrightarrow4m^2+8m+4-3m^2-24=0\)

\(\Leftrightarrow m^2+8m-20=0\)

Đến đây bạn tự tìm m là xong rồi

3.exiting

4.intesring

5.sad

6.intesring

7.same

8.same

1.

c, \(sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\\dfrac{\pi}{3}-x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\x=-\dfrac{2\pi}{3}+arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

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

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

1.

e, \(2sin2x+\sqrt{2}=0\)

\(\Leftrightarrow sin2x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x=sin\left(-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)

1.

f, \(sin\left(3x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=x-\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{4}=-x+\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{24}+k\pi\\x=\dfrac{11\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Đk:\(y^2-2x-5y+6\ge0\)

Pt (1)\(\Leftrightarrow\left(x^2-1\right)-\left(xy-y\right)+\left(x-1\right)=0\)

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

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

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

TH1: Thay x=1 vào pt (2) ta đc: \(3\sqrt{y^2-5y+4}=y+9\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+9\ge0\\9\left(x^2-5y+4\right)=y^2+18y+81\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge-9\\8y^2-63y-45=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{63+3\sqrt{601}}{16}\\y=\dfrac{63-3\sqrt{601}}{16}\end{matrix}\right.\) (tm)

TH2: Thay y=x+2 vào pt (2) ta đc:

\(\left(x-1\right)^2+3\sqrt{\left(x+2\right)^2-2x-5\left(x+2\right)+6}=x+2+9\)

\(\Leftrightarrow x^2-3x-10+3\sqrt{x^2-3x}=0\)

Đặt \(t=\sqrt{x^2-3x}\left(t\ge0\right)\)

Pttt: \(t^2-10+3t=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow2=\sqrt{x^2-3x}\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=6\\y=1\end{matrix}\right.\) (tm)

Vậy \(\left(x;y\right)=\text{​​}\left\{\left(1;\dfrac{63+3\sqrt{601}}{16}\right);\left(1;\dfrac{63-3\sqrt{601}}{16}\right),\left(4;6\right),\left(-1;1\right)\right\}\)

Xét pt đầu:

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

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

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

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

- Với \(x=1\) thay xuống pt dưới:

\(3\sqrt{y^2-5y+4}=y+9\) \(\left(y\ge-9\right)\)

\(\Leftrightarrow9\left(y^2-5y+4\right)=y^2+18y+81\)

\(\Leftrightarrow8y^2-63y-45=0\)

\(\Rightarrow y=\dfrac{63\pm3\sqrt{601}}{16}\) (thỏa mãn)

- Với \(y=x+2\) thay xuống pt dưới:

\(\left(x-1\right)^2+3\sqrt{x^2-3x}=x+11\) (ĐKXĐ: ....)

\(\Leftrightarrow x^2-3x+3\sqrt{x^2-3x}-10=0\)

Đặt \(\sqrt{x^2-3x}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x}=2\Leftrightarrow x^2-3x-4=0\)

\(\Leftrightarrow...\)

f: \(3ab-6a+b-2\)

\(=3a\left(b-2\right)+\left(b-2\right)\)

\(=\left(b-2\right)\left(3a+1\right)\)