A) \(A=-3x^2+x+1\)

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

\(A=-3\left(x^2-2\cdot\dfrac{1}{6}\cdot x+\dfrac{1}{36}-\dfrac{13}{36}\right)\)

\(A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)

Mà: \(-3\left(x-\dfrac{1}{6}\right)^2\le0\forall x\)

\(\Rightarrow A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\le\dfrac{13}{12}\forall x\)

Dấu "=" xảy ra khi:

\(x-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{6}\)

Vậy: \(A_{max}=\dfrac{13}{12}.khi.x=\dfrac{1}{6}\)

B) \(B=2x^2-8x+1\)

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

\(B=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)

\(B=2\left(x-2\right)^2-7\)

Mà: \(2\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow B=2\left(x-2\right)^2-7\ge-7\forall x\)

Dấu "=" xảy ra khi:

\(x-2=0\Rightarrow x=2\)

Vậy: \(B_{min}=2.khi.x=2\)

câu a) bạn viết sai đề rồi

d. Áp dụng BĐT Caushy Schwartz ta có:

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)

-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

c. Bạn kiểm tra lại đề nhé.

b. \(5x\left(2-x\right)=-5x\left(x-2\right)=-5\left(x^2-2x\right)=-5\left(x^2-2x+1-1\right)=-5\left(x-1\right)^2+5\le5\)-Dấu bằng xảy ra \(\Leftrightarrow x=1\)

a.

\(\left(80-2x\right)\left(50-2x\right)x=\dfrac{2}{3}\left(40-x\right)\left(50-2x\right)3x\le\dfrac{2}{3}\left(\dfrac{40-x+50-2x+3x}{3}\right)^3=18000\)

Dấu "=" xảy ra khi \(40-x=50-2x=3x\Leftrightarrow x=10\)

b.

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

Dấu "=" xảy ra khi \(x=2-x\Rightarrow x=1\)

c.

Biểu thức này chỉ có min, ko có max

d.

\(x+y\le1\Rightarrow-\left(x+y\right)\ge-1\)

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}=\left(4x+\dfrac{1}{x}\right)+\left(4y+\dfrac{1}{y}\right)-3\left(x+y\right)\ge2\sqrt{\dfrac{4x}{x}}+2\sqrt{\dfrac{4y}{y}}-3.1=5\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Bài 1:

Ta có: \(2x+\left|x-3\right|=4\)

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

Điều kiện: \(4-2x\ge0\Leftrightarrow2x\le4\Rightarrow x\le2\)

\(PT\Leftrightarrow\orbr{\begin{cases}x-3=4x-2\\x-3=2-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{3}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

Vậy x = 1

Bài 2:

a) Ta có: \(A=\left|3x+5\right|+4\ge4\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|3x+5\right|=0\Rightarrow x=-\frac{5}{3}\)

Vậy Min(A) = 4 khi x = -5/3

b) Ta có: \(B=-\left|2x+1\right|+10\le10\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|2x+1\right|=0\Rightarrow x=-\frac{1}{2}\)

Vậy Max(B) = 10 khi x = -1/2

Bài 1:

\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)

2:

a: =-(x^2-3x+1)

=-(x^2-3x+9/4-5/4)

=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn

b: =-2(x^2+3/2x+3/2)

=-2(x^2+2*x*3/4+9/16+15/16)

=-2(x+3/4)^2-15/8<0 với mọi x

Bài 1:

\(B=4+x^2+x=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\\ Vậy:B>0\forall x\in R\)

Bài 2: 

a: \(A=x^2+8x\)

\(=x^2+8x+16-16\)

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

Dấu '=' xảy ra khi x=-4

b: \(B=-2x^2+8x-15\)

\(=-2\left(x^2-4x+\dfrac{15}{2}\right)\)

\(=-2\left(x^2-4x+4+\dfrac{7}{2}\right)\)

\(=-2\left(x-2\right)^2-7\le-7\)

Dấu '=' xảy ra khi x=2

c: \(C=x^2-4x+7\)

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

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

Dấu '=' xảy ra khi x=2

e: \(E=x^2-6x+y^2-2y+12\)

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

\(=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=3 và y=1

\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)

Ta có \(64:59R5\Rightarrow64^n:59R5\)

Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)

Vậy \(A⋮59\)

(\(R\) là dư)

\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)