Trương Phạm Đăng Khoa ai xóa câu hỏi đi ấy

Vào trang của toi có câu trả lời :)

Bài 6:
a)Thay m=7 vào pt ta có:  \(x^2-8x+7=0\) \(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

b)Để pt có hai nghiệm pb \(\Leftrightarrow\Delta>0\) \(\Leftrightarrow64-4m>0\Leftrightarrow m< 16\)

Theo định lí viet có: \(\left\{{}\begin{matrix}x_1+x_2=8\\x_1x_2=m\end{matrix}\right.\)

kết hợp với giả thiết ta có hệ: \(\left\{{}\begin{matrix}x_1+x_2=8\\x_1=3x_2\\x_1x2=m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x_2+x_2=8\\x_1=3x_2\\x_1x_2=m\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_2=2\\x_1=6\\x_1x_2=m\end{matrix}\right.\)\(\Rightarrow m=2.6=12\) (tm)

Bài 7: 

Để pt \(x^2-2\left(m+1\right)x+m^2+m-1=0\) có hai nghiệm

\(\Leftrightarrow\Delta\ge0\) \(\Leftrightarrow4\left(m+1\right)^2-4\left(m^2+m-1\right)\ge0\) \(\Leftrightarrow8+4m\ge0\) \(\Leftrightarrow m\ge-2\)
Theo định lí viet có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+m-1\end{matrix}\right.\)

\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=4\left(m+1\right)^2-2\left(m^2+m-1\right)\)

\(=2m^2+6m+6=2\left(m^2+2.m.\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{3}{2}\)

\(=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\forall m\)

=>minA=\(\dfrac{3}{2}\Leftrightarrow m=-\dfrac{3}{2}\) (tm)

Có \(ab=\sqrt[3]{\left(7+\sqrt{50}\right)\left(7-\sqrt{50}\right)}\)\(=\sqrt[3]{7^2-\left(\sqrt{50}\right)^2}=-1\)

\(a^3+b^3=7+\sqrt{50}+7-\sqrt{50}=14\)

Lại có: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(\Leftrightarrow14=M^3+3M\) (do a+b=M)

\(\Leftrightarrow M^3+3M-14=0\)

\(\Leftrightarrow M^3-2M^2+2M^2-4M+7M-14=0\)

\(\Leftrightarrow M^2\left(M-2\right)+2M\left(M-2\right)+7\left(M-2\right)=0\)

\(\Leftrightarrow\left(M-2\right)\left(M^2+2M+7\right)=0\)

\(\Leftrightarrow M=2\) (do \(M^2+2M+7=\left(M+1\right)^2+6>0\forall M\))

=> M là số chẵn

\(N=a^7+b^7=\left(a^6+b^6\right)\left(a+b\right)-ab\left(a^5+b^5\right)\)

\(=2\left(a^6+b^6\right)-\left(-1\right)\left(a^5+b^5\right)\)

\(=2\left(a^6+b^6\right)+\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\)

\(=2\left(a^6+b^6\right)+14\left(a^2+b^2\right)-2a^2b^2\)

\(=2\left(a^6+b^6+7a^2+7b^2-a^2b^2\right)⋮2\)

=>N là số chẵn

\(\Delta ABC\sim\Delta DEF\)\(\Rightarrow\dfrac{AB}{DE}=\dfrac{BC}{EF}\)

\(\Leftrightarrow AB=\dfrac{BC.DE}{EF}=\dfrac{12.12}{9}=16\)

Ý B

Đk: \(x\ne2;y\ne\dfrac{1}{2}\)

Đặt \(a=\dfrac{1}{x-2},b=\dfrac{1}{2y-1}\) (a,b khác 0)

Có hệ: \(\left\{{}\begin{matrix}a+5b=3\\3a-b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\15a-5b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}16a=8\\3a-b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=3a-1=\dfrac{1}{2}\end{matrix}\right.\)(tm)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{1}{2}\\\dfrac{1}{2y-1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3}{2}\end{matrix}\right.\)(tm)

ý 5 ấy, bạn xem có đúng ko chứ toi cx ko bt đâu :<

\(\left\{{}\begin{matrix}x^2-xy-2y^2=0\\3x+y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\left(1-3x\right)-2\left(1-3x\right)^2=0\\y=1-3x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-14x^2+11x-2=0\\y=1-3x\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{7}\end{matrix}\right.\\y=1-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{7}\\y=\dfrac{1}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy...

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\) 

\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)

Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)

\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)

minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\)  \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)

maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)