Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

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

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

dài quá bạn ơi viết từng câu thôi

bạn xem lại đề bài 1 là GTNN hay GTLN nha

1) \(A=\frac{2018x^2-2.2018x+2018^2}{2018x^2}=\frac{\left(x-2018\right)^2+2017x^2}{2018x^2}=\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\)

vì \(\frac{\left(x-2018\right)^2}{2018x^2}\ge0\Rightarrow\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\ge\frac{2017}{2018}\)

dấu = xảy ra khi x-2018=0

=> x=2018

Vậy Min A=\(\frac{2017}{2017}\)khi x=2018

2) \(B=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}=1+\frac{10}{3.x^2+9x+7}\)

\(=1+\frac{10}{3.\left(x^2+9x\right)+7}=1+\frac{10}{3.\left[x^2+\frac{2.x.3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{4}+7}=1+\frac{10}{3.\left(x+\frac{9}{2}\right)^2+\frac{1}{4}}\)

để B lớn nhất => \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)nhỏ nhất

mà \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)vì \(3.\left(x+\frac{3}{2}\right)^2\ge0\)

dấu = xảy ra khi \(x+\frac{3}{2}=0\)

=> x=\(-\frac{3}{2}\)

Vậy maxB=\(41\)khi x=\(-\frac{3}{2}\)

3) \(M=\frac{3x^2+14}{x^2+4}=\frac{3.\left(x^2+4\right)+2}{x^2+4}=3+\frac{2}{x^2+4}\)

để M lớn nhất => x²+4 nhỏ nhất

mà \(x^2+4\ge4\)(vì x² lớn hơn hoặc bằng 0)

dấu = xảy ra khi x² =0

=> x=0

Vậy Max M\(=\frac{7}{2}\)khi x=0

ps: bài này khá dài, sai sót bỏ qua =))

ê viết lộn dòng này :v

\(MinA=\frac{2017}{2018}\)nha 

Thanks. <3

\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)

Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)

Dấu \("="\Leftrightarrow x=-5\)

đề dài v~

1.

a) \(f\left(x\right)=5x^2-2x+1\)

\(5f\left(x\right)=25x^2-10x+5\)

\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)

\(5f\left(x\right)=\left(5x-1\right)^2+4\)

Mà  \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow5f\left(x\right)\ge4\)

\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)

Dấu " = " xảy ra khi :

\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy ....

b)  \(P\left(x\right)=3x^2+x+7\)

\(3P\left(x\right)=9x^2+3x+21\)

\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)

\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)

Mà  \(\left(3x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)

\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)

Dấu "=" xảy ra khi :

\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)

Vậy ...

c)  \(Q\left(x\right)=5x^2-3x-3\)

\(5Q\left(x\right)=25x^2-15x-15\)

\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)

\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(5x-\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)

\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)

Dấu "=" xảy ra khi :

\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)

Vậy ...

2.

a)  \(f\left(x\right)=-3x^2+x-2\)

\(-3f\left(x\right)=9x^2-3x+6\)

\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)

\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)

Mà  \(\left(3x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)

\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)

Dấu "=" xảy ra khi :

\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

b)  \(P\left(x\right)=-x^2-7x+1\)

\(-P\left(x\right)=x^2+7x-1\)

\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)

\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x+\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)

\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)

Vậy ...

c)  \(Q\left(x\right)=-2x^2+x-8\)

\(-2Q\left(x\right)=4x^2-2x+16\)

\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)

\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)

Mà :  \(\left(2x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)

\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)

Dấu "=" xảy ra khi :

\(2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy ...

3/

Ta có x + y + z = 0

<=> (x + y + z)² = 0

<=> x² + y² + z² + 2xy + 2xz + 2yz = 0

<=> x² + y² + z² + 2(xy + xz + yz) = 0

Thay xy + yz + xz = 0 vào BT trên

=> x² + y² + z² = 0

Mà \(x^2\ge0;y^2\ge0;z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)

=> x = y = z (đpcm)

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

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

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

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

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)