a)A=4(x+11/8)^2 -153/16

Min A=-153/16 khi x=-11/8

b)B=3(x-1/3)^2 -4/3

Min B=-4/3 khi x=1/3

Bài 1:

a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)

\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)

b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)

\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)

Bài 2:

a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)

b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)

\(maxB=11\Leftrightarrow x=-2\)

Bài 1: 

a: Ta có: \(A=4x^2+11x-2\)

\(=4\left(x^2+\dfrac{11}{4}x-\dfrac{1}{2}\right)\)

\(=4\left(x^2+2\cdot x\cdot\dfrac{11}{8}+\dfrac{121}{64}-\dfrac{153}{64}\right)\)

\(=4\left(x+\dfrac{11}{8}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{11}{8}\)

b: Ta có: \(B=3x^2-2x-1\)

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

\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{4}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

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

\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)

\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)

\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)

\(\Rightarrow4x=21\)

\(\Rightarrow x=\dfrac{21}{5}\)

Bài 2: 

a: Ta có: \(x^2+4x+7\)

\(=x^2+4x+4+3\)

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

phân tích các số sau ra thừa số nguyên tố 48;56;84;105;360

Để (d) và (d') song song 

khi 2m-1 = 3; và 1 khác -2 (luôn đúng)

<=> 2m=4 <=> m=2 (thỏa mãn)

Vậy m=2 thì (d) // (d')

A(x)+B(x)-C(x)

=x^3+2x^2+3x+1-x^3+x+1-2x^2+1=0

=>4x+3=0

=>x=-3/4

b) Theo đề ra, ta có:

 \(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{6}\)

\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{6}=\frac{z}{5}\)

Áp dụng tính chất của dãy tỷ số bằng nhau

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{5}=\frac{x-y}{10-6}=1\)

\(\Rightarrow x=1.10=10\)

\(\Rightarrow y=1.6\)

\(\Rightarrow z=1.5=5\)

a: Để hai đồ thị song song thì \(\left\{{}\begin{matrix}m^2-1=3\\m-3\ne2m-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\-m\ne2\end{matrix}\right.\Leftrightarrow m=2\)

b: Để hai đồ thị trùng nhau thì \(\left\{{}\begin{matrix}m^2-1=3\\m-3=2m-1\end{matrix}\right.\)

hay m=-2