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31/

\(3z^2-2z+27=0\)

\(\Delta'=\left(-1\right)^2-3.27=1-3.27=-80\)

\(\Delta'\) có 2 căn bậc 2 là \(\pm4i\sqrt{5}\)

\(\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1+4i\sqrt{5}}{3}\\z_2=\dfrac{1-4i\sqrt{5}}{3}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\left(\dfrac{1}{3}\right)^2+\left(\dfrac{4\sqrt{5}}{3}\right)^2}=3\)

\(\Rightarrow z_1\left|z_2\right|+z_2\left|z_1\right|=1+4i\sqrt{5}+1-4i\sqrt{5}=2\) => A

32/ \(\Delta'=4-29=-25\Rightarrow\left\{{}\begin{matrix}z_1=-2+5i\\z_2=-2-5i\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{2^2+5^2}=\sqrt{29}\)

\(\Rightarrow\left|z_1\right|^4+\left|z_2\right|^4=2.\sqrt{29^4}=1682\) => B

33/ \(\Delta=1-12=-11\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1+i\sqrt{11}}{6}\\z_2=\dfrac{1-i\sqrt{11}}{6}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\left(\dfrac{1}{6}\right)^2+\left(\dfrac{\sqrt{11}}{6}\right)^2}=\dfrac{\sqrt{3}}{3}\)

\(\Rightarrow\left|z_1\right|+\left|z_2\right|=\dfrac{2\sqrt{3}}{3}\) => D

34/ \(\Delta=1-4.3.2=-23\Rightarrow\left\{{}\begin{matrix}z_1=\dfrac{1-i\sqrt{23}}{6}\\z_2=\dfrac{1+i\sqrt{23}}{6}\end{matrix}\right.\Rightarrow\left|z_1\right|=\left|z_2\right|=\sqrt{\dfrac{1}{36}+\dfrac{23}{36}}=\dfrac{\sqrt{6}}{3}\)

\(\Rightarrow T=2.\left(\dfrac{\sqrt{6}}{3}\right)^2=\dfrac{4}{3}\) => C

Tks bn nhiều

\(y'=\dfrac{\left(-2x+2\right)\left(x-3\right)-\left(-x^2+2x+c\right)}{\left(x-3\right)^2}=\dfrac{-x^2+6x-6-c}{\left(x-3\right)^2}\)

\(\Rightarrow\) Cực đại và cực tiểu của hàm là nghiệm của: \(-x^2+6x-6-c=0\) (1)

\(\Delta'=9-\left(6+c\right)>0\Rightarrow c< 3\)

Gọi \(x_1;x_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}-x_1^2+6x_1-6=c\\-x_2^2+6x_2-6=c\end{matrix}\right.\)

\(\Rightarrow m-M=\dfrac{-x_1^2+2x_1+c}{x_1-3}-\dfrac{-x_2^2+2x_2+c}{x_2-3}=4\)

\(\Leftrightarrow\dfrac{-2x_1^2+8x_1-6}{x_1-3}-\dfrac{-2x_2^2+8x_2-6}{x_2-3}=4\)

\(\Leftrightarrow2\left(1-x_1\right)-2\left(1-x_2\right)=4\)

\(\Leftrightarrow x_2-x_1=2\)

Kết hợp với Viet: \(\left\{{}\begin{matrix}x_2-x_1=2\\x_1+x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2\\x_2=4\end{matrix}\right.\)

\(\Rightarrow c=2\)

Có 1 giá trị nguyên

Xét \(I_1=2\int\limits^{\dfrac{\pi}{2}}_0f\left(sinx\right)cosxdx=2\int\limits^{\dfrac{\pi}{2}}_0f\left(sinx\right)d\left(sinx\right)\)

Đặt \(sinx=t\Rightarrow t\in\left[0;1\right]\Rightarrow f\left(t\right)=5-t\)

\(I_1=2\int\limits^1_0\left(5-t\right)dt=9\)

Xết \(I_2=3\int\limits^1_0f\left(3-2x\right)dx=-\dfrac{3}{2}\int\limits^1_0f\left(3-2x\right)d\left(3-2x\right)\)

Đặt \(3-2x=t\Rightarrow t\in\left[1;3\right]\Rightarrow f\left(t\right)=t^2+3\)

\(I_2=-\dfrac{3}{2}\int\limits^1_3\left(t^2+3\right)dt=\dfrac{3}{2}\int\limits^3_1\left(t^2+3\right)dt=22\)

\(\Rightarrow I=9+22=31\)

Câu 35 Diện tích tam giác ABC là \(\dfrac{1}{2}\).AB.AC=\(\dfrac{1}{2}\).a\(^2\)

thể tích lăng trụ ABC.A'B'C'=a\(\sqrt{3}\).\(\dfrac{1}{2}\)a\(^2\)=a\(^3\)\(\dfrac{\sqrt{3}}{2}\)

Câu 36 Gọi O là giao điểm của AC và DB

AC=\(\sqrt{2}\)a => AO=\(\dfrac{\sqrt{2}}{2}\)a
Mà góc SAC=45 => SO=AO=\(\dfrac{\sqrt{2}}{2}\)a

thể tích khối chóp S.ABCD = \(\dfrac{1}{3}\).\(\dfrac{\sqrt{2}}{2}\)a.a\(^2\)=\(\dfrac{\sqrt{2}}{6}\)a\(^3\)

Câu 37 \(\dfrac{Vs.A'B'C'}{Vs.ABC}\)=\(\dfrac{SA'}{SA}\).\(\dfrac{SB'}{SB}\).\(\dfrac{SC'}{SC}\)=\(\dfrac{1}{8}\) 

tương tự \(\dfrac{Vs.D'B'C'}{Vs.DBC}\)=\(\dfrac{1}{8}\)=> \(\dfrac{Vs.A'B'C'D'}{Vs.ABCD}\)=\(\dfrac{1}{8}\)

Chọn A

Chọn A