Violympic toán 8

\(B=-2x^2-x+5\)

\(=-2\left(x^2+\dfrac{1}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}-\dfrac{41}{16}\right)\)

\(=-2\left(x+\dfrac{1}{4}\right)^2+\dfrac{41}{8}\le\dfrac{41}{8}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{4}=0\Leftrightarrow x=-\dfrac{1}{4}\)

Vậy Max B là : \(\dfrac{41}{8}\Leftrightarrow x=-\dfrac{1}{4}\)

\(A=-3x^2+x-2\)

\(=-3\left(x^2-\dfrac{1}{3}x+\dfrac{2}{3}\right)\)

\(=-3\left(x^2-2x.\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)

\(=-3\left[\left(x-\dfrac{1}{6}\right)^2+\dfrac{23}{36}\right]\)

\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{69}{26}\le-\dfrac{69}{26}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{1}{6}\)

Vậy Max A là : \(\dfrac{-69}{26}\Leftrightarrow x=\dfrac{1}{6}\)

\(B=-2x^2-x+5\)

\(=-2\left(x^2-\dfrac{1}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2-2x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{41}{16}\right)\)

\(=-2\left[\left(x-\dfrac{1}{4}\right)^2-\dfrac{41}{16}\right]\)

\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{41}{8}\le\dfrac{41}{8}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{4}\)

Vậy Max B là : \(\dfrac{41}{8}\Leftrightarrow x=\dfrac{1}{4}\)

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

\(=-x^2-2x-1-4x^2+12x-9\)

\(=-5x^2+10x-10\)

\(=-5\left(x^2-2x+1+1\right)\)

\(=-5\left[\left(x-1\right)^2+1\right]\)

\(=-5\left(x-1\right)^2-5\le-5\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Max C là : \(-5\Leftrightarrow x=1\)

\(E=2-5x^2-y^2-4xy+2x\)

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

\(=-\left(2x+y\right)^2-\left(x-1\right)^2+3\le3\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=1\end{matrix}\right.\)

Vậy Max E là : \(3\Leftrightarrow x=1;y=-2\)

\(B=x^2-x+5=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+5\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)(dấu "='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2})\)

Vậy \(B_{min}=\dfrac{19}{4}\) tại \(x=\dfrac{1}{2}\)

\(C=2x^2-x-1=2\left(x^2-\dfrac{1}{2}x-\dfrac{1}{2}\right)\)

\(=2\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)\)

\(=2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{16}\right]\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\) (dấu "='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2})\)

Vậy \(C_{min}=\dfrac{7}{8}\) tại \(x=\dfrac{1}{2}\)

\(D=3x^2+x+1=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}\right)\)

\(=3\left(x^2+2.x.\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}+\dfrac{1}{3}\right)\)

\(=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{36}\right]\)

\(=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{12}\ge\dfrac{11}{12}\) (dấu "='' xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\))

Vậy \(D_{min}=\dfrac{11}{12}\) tại \(x=-\dfrac{1}{6}\)

Tham khảo cách làm lớp 8

Câu hỏi của Nguyễn Đăng Khoa - Toán lớp 7 | Học trực tuyến

Tham khảo cách làm lớp 7

Câu hỏi của Nguyễn Thị Ngọc Lan - Toán lớp 7 - Học toán với OnlineMath

\(P=\left(1+2+3+...+n\right)^2\)

Ta có :

+) \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow1+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow2\left(ab+bc+ac\right)=-1\)

\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)

+) \(ab+bc+ac=-\dfrac{1}{2}\)

\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\dfrac{1}{2}\right)^2\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc.0=\dfrac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+0=\dfrac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)

+) \(a^2+b^2+c^2=1\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=1\)

\(\Rightarrow a^4+b^4+c^4+2.\dfrac{1}{4}=1\)

\(\Rightarrow a^4+b^4+c^4+\dfrac{1}{2}=1\)

\(\Rightarrow a^4+b^4+c^4=\dfrac{1}{2}\)

Vậy \(a^4+b^4+c^4=\dfrac{1}{2}\)

a: Xét ΔBAC có BA=BC

nên ΔBAC cân tại B

=>góc BAC=góc BCA

=>góc BCA=góc ACD

=>góc ACD=1/2góc ADC

góc ACD=1/3x90=30 độ

góc ADC=90-30=60 độ

=>góc BCD=60 độ; góc BAD=góc ABC=120 độ

b: Gọi Mlà trung điểm của CD

Vì ΔADC vuông tại A

mà M là trung điểm của CD

nên AM=MD=MC

Xét ΔMAD có MA=MD và góc D=60 độ

nên ΔMAD đều

=>AD=DM

=>AB=DM=MC

=>AB=1/2DC

\(=3x^4-6x^3-15x^2+x^3-2x^2-5x-x^2+2x+5\)

\(=\left(x^2-2x-5\right)\left(3x^2+x-1\right)\)