\(\left|x-y-5\right|+2007\left(y-3\right)^{2018}=0\)

Ta có: \(\left|x-y-5\right|+2007\left(y-3\right)^{2018}\ge0\forall x,y\)

\(\Rightarrow\) Để bt = 0 thì: \(\left\{{}\begin{matrix}\left|x-y-5=0\right|\\2007\left(y-3\right)^{2018}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=5\\y-3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

\(A=x^2-3x+1=x^2-2\cdot x\cdot1,5+1,5^2-1,25=\left(x-1,5\right)^2-1,25\ge-1,25\)

Vậy \(A_{min}=-1,25\) khi x - 1,5 = 0 \(\Rightarrow x=1,5\)

Câu 6,7,9: Ptđt thành nhân tử rồi tìm x, hoặc tính nhé! (3 bài này dụng phương pháp đặt nhân tử chung , dùng hđt , or nhóm nhé!)

Bài 8:

a) \(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

=> \(A_{min}=6\) khi \(x-2=0\Rightarrow x=2\)

b) \(B=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)

=> \(B_{min}=4\) khi \(x-1=0\Rightarrow x=1\)

8 mới đúng

tớ là cao thủ violympic nhưng ko thi đc huhu

a) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(\Rightarrow7x-7=6x+30\Rightarrow7x-6x=30+7\)

\(\Rightarrow x=37\)

b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy...............

Theo đề ta có: \(\dfrac{x}{1}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{y+z}{\dfrac{5}{2}}\)

và x + y + z = 280

Áp dụng t/c của dãy tỉ số bằng nhau có:

\(\dfrac{x}{1}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{y+z}{\dfrac{5}{2}}=\dfrac{x+y+y+z}{1+\dfrac{1}{4}+\dfrac{5}{2}}=\dfrac{280+y}{3,75}\)

\(\Rightarrow\dfrac{y}{\dfrac{1}{4}}=\dfrac{280+y}{3,75}\Rightarrow3,75y=\dfrac{1}{4}\left(280+y\right)\)

\(\Rightarrow3,75y=70+\dfrac{1}{4}y\Rightarrow3,75y-\dfrac{1}{4}y=70\)

\(\Rightarrow3,5y=70\Rightarrow y=\dfrac{70}{3,5}=20\)

Có: \(\dfrac{x}{1}=\dfrac{y}{\dfrac{1}{4}}\Rightarrow\dfrac{x}{1}=\dfrac{20}{\dfrac{1}{4}}\Rightarrow\dfrac{1}{4}x=20\Rightarrow x=20:\dfrac{1}{4}=80\)

\(\Rightarrow z=280-\left(x+y\right)=280-100=180\)

Vậy x = 80; y = 20; z = 180

\(x^3-8x^2+9x-2=x^3-7x^2-x^2+2x+7x-2=\left(x^3-7x^2+2x\right)-\left(x^2-7x+2\right)=x\left(x^2-7x+2\right)-\left(x^2-7x+2\right)=\left(x^2-7x+2\right)\left(x-1\right)\)

\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{3}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x+y=0\Rightarrow y=-\dfrac{1}{3}\end{matrix}\right.\)

3)

a) \(x\left(x-2\right)+x-2=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)