Câu 1:
TXĐ:D=R

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

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

=>f(x) là hàm số chẵn

Câu 9: B

Câu 10: A

Câu 1: C

Câu 2: A

Ai giải giúp em vs

Câu 15: 

\(A\cap B\)=[0;4)

\(A\cup B\)=(-3;5]

A\B=(-3;0)

Gọi G là trọng tâm tam giác ABC

\(\Rightarrow T=\sum\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\cdot\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)=3MG^2+\dfrac{4}{9}\cdot\left(m_a^2+m_b^2+m_c^2\right)=3MG^2+\dfrac{4}{9}\cdot\left(\dfrac{2b^2+2c^2-a^2}{4}+\dfrac{2a^2+2c^2-b^2}{4}+\dfrac{2b^2+2a^2-c^2}{4}\right)\) = \(3MG^2+\dfrac{1}{3}\left(a^2+b^2+c^2\right)\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)\) Dấu = xảy ra \(\Leftrightarrow M\equiv G\)

b) \(\dfrac{3\pi}{2}< \alpha< 2\pi\)\(\Rightarrow cos\alpha>0;sin\alpha< 0\)

Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)\(\Rightarrow cos\alpha=\dfrac{4}{5}\)

\(sin\alpha=-\sqrt{1-cos^2\alpha}=-\dfrac{3}{5}\)

\(sin\left(\alpha-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sin\alpha-cos\alpha\right)=\dfrac{\sqrt{2}}{2}\left(-\dfrac{3}{5}-\dfrac{4}{5}\right)=-\dfrac{7\sqrt{2}}{10}\)

Bài 2:

a) Gọi đt d vuông góc với đường thẳng \(\Delta\)có dạng: \(d:-4x+3y+c=0\)

\(A\in\left(d\right)\Rightarrow-4+3+c=0\Leftrightarrow c=1\)

Vậy \(d:-4x+3y+1=0\)

b) Gọi pt đường tròn (C) tâm A có dạng \(\left(C\right):\left(x-1\right)^2+\left(y-1\right)^2=R^2\)

Vì (C) tiếp xúc với \(\Delta\)

\(\Rightarrow\)\(R=d_{\left(A;\Delta\right)}=\dfrac{\left|3+4+5\right|}{\sqrt{3^2+4^2}}=\dfrac{12}{5}\)

\(\Rightarrow\left(C\right):\left(x-1\right)^2+\left(y-1\right)^2=\dfrac{144}{25}\)

Vậy...

6.

\(\Leftrightarrow x^2+4x+3>m\) ; \(\forall x>1\)

\(\Leftrightarrow m< \min\limits_{x>1}\left(x^2+4x+3\right)\)

Xét hàm \(f\left(x\right)=x^2+4x+3\) với \(x>1\)

\(-\dfrac{b}{2a}=-2< 1\) ; \(f\left(1\right)=8\Rightarrow f\left(x\right)>8\) ; \(\forall x>1\)

\(\Rightarrow m\le8\)

7.

Do C thuộc d nên tọa độ có dạng: \(C\left(-2c-1;c\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;6\right)\\\overrightarrow{CA}=\left(2c;1-c\right)\end{matrix}\right.\)

\(AB\perp AC\Leftrightarrow\overrightarrow{AB}.\overrightarrow{AC}=0\Leftrightarrow4.2c+4\left(1-c\right)=0\)

\(\Leftrightarrow4c+4=0\Rightarrow c=-1\Rightarrow C\left(1;-1\right)\)

b.

 \(AB=\sqrt{4^2+6^2}=2\sqrt{13}\)

Phương trình đường thẳng AB qua A và nhận \(\left(3;-2\right)\) là 1 vtpt có dạng:

\(3\left(x+1\right)-2\left(y-1\right)=0\Leftrightarrow3x-2y+5=0\)

Do d thuộc d nên tọa độ có dạng: \(D\left(-2d-1;d\right)\)

\(S_{ABD}=\dfrac{1}{2}AB.d\left(D;AB\right)=50\)

\(\Leftrightarrow\dfrac{\sqrt{13}\left|3\left(-2d-1\right)-2d+5\right|}{\sqrt{3^2+\left(-2\right)^2}}=50\)

\(\Leftrightarrow\left|-8d+2\right|=50\Rightarrow\left[{}\begin{matrix}d=-6\\d=\dfrac{13}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}D\left(11;-6\right)\\D\left(-14;\dfrac{13}{2}\right)\end{matrix}\right.\)

2.

a, Gọi \(C=\left(-2m-1;m\right)\) là điểm cần tìm

\(AB=2\sqrt{13};AC=\sqrt{5m^2-2m+1};BC=\sqrt{5m^2+2m+65}\)

Ta có \(BC^2=AB^2+AC^2\)

\(\Leftrightarrow5m^2+2m+65=52+5m^2-2m+1\)

\(\Leftrightarrow m=-3\)

\(\Rightarrow C=\left(5;-3\right)\)

b, Gọi \(D=\left(-2n-1;n\right)\) là điểm cần tìm

Đường thẳng AB có phương trình \(\dfrac{x+1}{4}=\dfrac{y-1}{6}\Leftrightarrow3x-2y+5=0\)

Khoảng cách từ \(D\) đến \(AB\):

\(d\left(D;AB\right)=\dfrac{\left|3\left(-2n-1\right)-2n+5\right|}{\sqrt{3^2+2^2}}=\dfrac{\left|-8n+2\right|}{\sqrt{13}}\)

\(S_{ABC}=\dfrac{1}{2}.\dfrac{\left|-8n+2\right|}{\sqrt{13}}.2\sqrt{13}=50\)

\(\Rightarrow\left|4n-1\right|=25\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-6\\n=\dfrac{13}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}N=\left(11;-6\right)\\N=\left(-14;\dfrac{13}{2}\right)\end{matrix}\right.\)

9C

7D

6D

8D

11A

10C

13B

đk: \(\left\{{}\begin{matrix}x\ge-1\\y\ge-2\end{matrix}\right.\)

TheoBĐT Bunhiacopxki ,ta có: \(x-3\sqrt{x+1}=3\sqrt{y+2}-y\)

\(\Rightarrow\left(x+y\right)^2-9\left(\sqrt{x+1}+\sqrt{y+2}\right)^2\le9.2\left(x+y+3\right)\)

\(\Leftrightarrow\left(x+y\right)^2-18\left(x+y\right)-54\le0\)

\(\Rightarrow x+y\le9+3\sqrt{15}\Rightarrow P\le9+3\sqrt{15}\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+y=9+3\sqrt{15}\\\sqrt{x+1}=\sqrt{y+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+3\sqrt{15}}{2}\\y=\dfrac{8+3\sqrt{15}}{2}\end{matrix}\right.\)

Vậy Max P = \(9+3\sqrt{15}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10+3\sqrt{15}}{2}\\y=\dfrac{8+3\sqrt{15}}{2}\end{matrix}\right.\)

===> Chọn D