\(A=x^2-4x-10\)

\(A=x^2-4x+4-14\)

\(A=\left(x^2-2.x.2+2^2\right)-14\)

\(A=\left(x-2\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\)

Nên \(\left(x-2\right)^2-14\ge-14\)

Vậy GTNN của \(A=-14\) khi \(x-2=0\Leftrightarrow x=2\)

\(x^3-x+6=0\)

\(\Leftrightarrow x^3-4x+3x+6=0\)

\(\Leftrightarrow\left(x^3-4x\right)+\left(3x+6\right)=0\)

\(\Leftrightarrow x\left(x^2-4\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow x\left(x^2-2^2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-2\right)+3\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-2x+3=0\left(loai\right)\end{matrix}\right.\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

Số lượng số hạng của dãy số cách đều \(\text{Số số hạng}=\left(\text{số hạng cuối}-\text{số hạng đầu}\right):d+1\)(trong đó d là khoảng cách giữa hai số hạng liên tiếp)

\(\left(99-1\right):1+1=98+1=99\)

Tổng của dãy số B là số cách đều nên ta có \(\text{tổng}=\dfrac{\left(\text{Số đầu}+\text{Số cuối}\right)\times\text{Số lượng số hạng}}{2}\)

\(B=\dfrac{\left(1+99\right).99}{2}=\dfrac{9900}{2}=4950\)

Vậy B=4950

\(x^3-x+6=0\)

\(\Leftrightarrow x^3-4x-3x+6=0\)

\(\Leftrightarrow\left(x^3-4x\right)-\left(3x-6\right)=0\)

\(\Leftrightarrow x\left(x^2-4\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x^2-2^2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x+2\right)-3\right]\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-3\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-x+3x-3\right)\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)+3\left(x-1\right)\right]\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x+3=0\end{matrix}\right.\)

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

Vậy \(x=2\) ; \(x=1\) ; \(x=-3\)

\(3x=y;5y=4z\Rightarrow\dfrac{x}{1}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x}{4}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{6x}{24}=\dfrac{7y}{84}=\dfrac{8z}{120}=\dfrac{6x+7y+8z}{24+84+120}=\dfrac{456}{228}=2\)

\(\dfrac{6x}{24}=2\Rightarrow6x=24.2=48\Rightarrow x=\dfrac{48}{6}=8\)

\(\dfrac{7y}{84}=2\Rightarrow7y=84.2=168\Rightarrow y=\dfrac{168}{7}=24\)

\(\dfrac{8z}{120}=2\Rightarrow8z=120.2=240\Rightarrow z=\dfrac{240}{8}=30\)

Vậy x=8 ; y=24 ; z=30

Hoàng Thị Ngọc Anh

từ \(15a=10b=6c\Rightarrow\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a-b+c}{\dfrac{1}{15}-\dfrac{1}{10}+\dfrac{1}{6}}=\dfrac{-24}{\dfrac{2}{15}}=-180\)

\(\dfrac{a}{\dfrac{1}{15}}=-180\Rightarrow a=\dfrac{1}{15}.-180=-12\)

\(\dfrac{b}{\dfrac{1}{10}}=-180\Rightarrow b=\dfrac{1}{10}.-180=-18\)

\(\dfrac{c}{\dfrac{1}{6}}=-180\Rightarrow c=\dfrac{1}{6}.-180=-30\)

Vậy \(a=-12\); \(b=-18\); \(c=-30\)

84+65+54+76

=152+130

=282

a)\(A=4x^2+4x+11\)

\(\Leftrightarrow A=4x^2+4x+1+10\)

\(\Leftrightarrow A=\left(2x+1\right)^2+10\)

Vì \(\left(2x+1\right)^2\ge0\)

Nên \(\left(2x+1\right)^2+10\ge10\)

Vậy GTNN của A=10 khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)

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

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

\(\Leftrightarrow B=-2x^2+2x-\dfrac{1}{2}-\dfrac{9}{2}\)

\(\Leftrightarrow B=-\left(2x^2-2x+\dfrac{1}{2}\right)-\dfrac{9}{2}\)

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

\(\Leftrightarrow B=-2\left(x^2-2.x\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{9}{2}\)

\(\Leftrightarrow B=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

Do đó \(-\left(x-\dfrac{1}{2}\right)^2\le0\)

Nên \(-\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)

Vậy GTLN của \(B=\dfrac{-9}{2}\) khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=4x^2-12x\)

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

\(\Leftrightarrow C=\left(4x^2-12x+9\right)-9\)

\(\Leftrightarrow C=\left(2x-3\right)^2-9\)

Vì \(\left(2x-3\right)^2\ge0\)

Nên \(\left(2x-3\right)^2-9\ge-9\)

Vậy GTNN của \(C=-9\) khi \(2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)

d) \(D=5-x^2+2x-4y^2-4y\)

\(\Leftrightarrow D=7-1-1-x^2+2x-4y^2-4y\)

\(\Leftrightarrow D=-x^2+2x-1-4y^2-4y-1+7\)

\(\Leftrightarrow D=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\)

\(\Leftrightarrow D=-\left(x-1\right)^2-\left(2y+1\right)^2+7\)

Vậy GTLN của \(D=7\) khi \(\left\{{}\begin{matrix}x-1=0\Leftrightarrow x=1\\2y+1=0\Leftrightarrow y=\dfrac{-1}{2}\end{matrix}\right.\)