3) Ta có: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m^2-6\right)\)

\(=\left(2m-2\right)^2-4\left(m^2-6\right)\)

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

\(=-8m+28\)

Để phương trình có hai nghiệm phân biệt x1;x2 thì Δ>0

\(\Leftrightarrow-8m+28>0\)

\(\Leftrightarrow-8m>-28\)

hay \(m< \dfrac{7}{2}\)

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{1}=2m-2\\x_1x_2=m^2-6\end{matrix}\right.\)

Ta có: \(x_1^2+x_2^2=16\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=16\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(m^2-6\right)-16=0\)

\(\Leftrightarrow4m^2-8m+4-2m^2+12-16=0\)

\(\Leftrightarrow2m^2-8m=0\)

\(\Leftrightarrow2m\left(m-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\left(nhận\right)\\m=4\left(loại\right)\end{matrix}\right.\)

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

Xét ΔBAC có \(cosB=\frac{a^2+c^2-b^2}{2\cdot a\cdot c}\)

=>\(\left(4\sqrt2\right)^2+10^2-b^2=2\cdot4\sqrt2\cdot10\cdot cos45=8\sqrt2\cdot10\cdot\frac{\sqrt2}{2}=80\)

=>\(b^2=32+100-80=32+20=52\)

=>\(b=\sqrt{52}=2\sqrt{13}\)

Xét ΔABC có cos C=\(\frac{a^2+b^2-c^2}{2\cdot a\cdot b}\)

=>cosC=\(\frac{32+52-100}{2\cdot4\sqrt2\cdot2\sqrt{13}}=\frac{-16}{16\sqrt{26}}=-\frac{1}{\sqrt{26}}\)

=>\(\sin C=\sqrt{1-cos^2C}=\frac{5}{\sqrt{26}}\)

Diện tích tam giác CAB là:

\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)

\(=\frac12\cdot\frac{5}{\sqrt{26}}\cdot2\sqrt{13}\cdot4\sqrt2=\frac{5\cdot2\cdot4}{2}=5\cdot4=20\)

Xét ΔABC có \(\frac{AB}{\sin C}=2R\)

=>\(2R=10:\frac{5}{\sqrt{26}}=\frac{10\sqrt{26}}{5}=2\sqrt{26}\)

=>\(R=\sqrt{26}\)

Ta có: \(S_{BCA}=\frac12\cdot AB\cdot AC\cdot\sin A\)

=>\(\frac12\cdot10\cdot2\sqrt{13}\cdot\sin A=20\)

=>\(\sin A=\frac{20}{10\sqrt{13}}=\frac{2}{\sqrt{13}}\)

\(S_{ACB}=\frac12\cdot BC\cdot h_{A}\)

=>\(\frac12\cdot4\sqrt2\cdot h_{A}=20\)

=>\(h_{A}=\frac{20}{2\sqrt2}=\frac{10}{\sqrt2}=5\sqrt2\)

A=|x+1,5|-4>=-4

-Vậy: MIN A=-4 tại x+1,5=0=>x=-1,5

B=|x+1|+|y-1|+2>=2

-Vậy: MIN B=2 tại x=-1;y=1

Xét ΔBAC có \(cosB=\frac{a^2+c^2-b^2}{2\cdot a\cdot c}\)

=>\(\left(4\sqrt2\right)^2+10^2-b^2=2\cdot4\sqrt2\cdot10\cdot cos45=8\sqrt2\cdot10\cdot\frac{\sqrt2}{2}=80\)

=>\(b^2=32+100-80=32+20=52\)

=>\(b=\sqrt{52}=2\sqrt{13}\)

Xét ΔABC có cos C=\(\frac{a^2+b^2-c^2}{2\cdot a\cdot b}\)

=>cosC=\(\frac{32+52-100}{2\cdot4\sqrt2\cdot2\sqrt{13}}=\frac{-16}{16\sqrt{26}}=-\frac{1}{\sqrt{26}}\)

=>\(\sin C=\sqrt{1-cos^2C}=\frac{5}{\sqrt{26}}\)

Diện tích tam giác CAB là:

\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)

\(=\frac12\cdot\frac{5}{\sqrt{26}}\cdot2\sqrt{13}\cdot4\sqrt2=\frac{5\cdot2\cdot4}{2}=5\cdot4=20\)

Xét ΔABC có \(\frac{AB}{\sin C}=2R\)

=>\(2R=10:\frac{5}{\sqrt{26}}=\frac{10\sqrt{26}}{5}=2\sqrt{26}\)

=>\(R=\sqrt{26}\)

Ta có: \(S_{BCA}=\frac12\cdot AB\cdot AC\cdot\sin A\)

=>\(\frac12\cdot10\cdot2\sqrt{13}\cdot\sin A=20\)

=>\(\sin A=\frac{20}{10\sqrt{13}}=\frac{2}{\sqrt{13}}\)

\(S_{ACB}=\frac12\cdot BC\cdot h_{A}\)

=>\(\frac12\cdot4\sqrt2\cdot h_{A}=20\)

=>\(h_{A}=\frac{20}{2\sqrt2}=\frac{10}{\sqrt2}=5\sqrt2\)

khó quá

anh

sai đề bạn ạ

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-3\right)^2-4\left(-2\right)\left(-m+1\right)>0\\x_1+x_2=\dfrac{3}{-2}< 0\\x_1x_2=\dfrac{-m+1}{-2}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17-8m>0\\-m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{17}{8}\\m>1\end{matrix}\right.\Leftrightarrow1< m< \dfrac{17}{8}\)

\(2,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-4\right)^2-4\left(-3\right)\left(-2m+1\right)\ge0\\x_1+x_2=\dfrac{4}{-3}< 0\\x_1x_2=\dfrac{-2m+1}{-3}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28-24m\ge0\\-2m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{7}{6}\\m>\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< m\le\dfrac{7}{6}\)

Giúp em câu e bài 1,bài 2,3 với 

1 C

2 C

3 A

4 B

5 C

6 A

7 C

8 B

9 D

Em cảm ơn ạ