\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)

\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)

\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)

\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)

\("="\Leftrightarrow x=3\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\)

=\(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge6\)(1)

mặt khác 5-2x-x²=6-(x+1)²\(\le6\)(2)

từ (1) và (2)=>dấu = xảy ra khi VP =6 =VTtức x=-1

b)\(\sqrt{3x^2+6x+12}\)+\(\sqrt{5x^4+10x^2+9}\)

=\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2+1\right)^2+4}>5\)(x²+1>0)(1')

mặt khác VP=5-2(x+1)²\(\le\)5(2')

từ (1') và (2')=> pt vô nghiệm

M = x⁴ - 6x³ + 10x² - 6x + 9

M = (x² - 6x + 9) + x⁴ - 6x³ + 9x²

M = (x - 3)² + x²(x² - 6x + 9)

M = (x - 3)².(1 + x²)

Ta có:\(\left(x-3\right)^2\ge0;\left(1+x^2\right)\ge1\)

\(\Rightarrow M\ge1\)

Dấu 'x' xảy ra khi:

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

Vậy Mmin = 1 khi x = 3

Chúc bạn học tốt!!!

Mình giải lại từ dòng số 6 nhé!!!

=> M = 0 

Dấu '=' xảy ra khi:

(x - 3)² = 0 => x - 3 = 0

=> x = 3

Vậy Mmin = 0 khi x = 3

\(A=\dfrac{x}{\left(x+2022\right)^2}=\dfrac{x}{x^2+4044x+2022^2}=\dfrac{1}{x+4044+\dfrac{2022^2}{x}}=\dfrac{1}{\left(x+\dfrac{2022^2}{x}\right)+4044}\le\dfrac{1}{2.\sqrt{x}.\sqrt{\dfrac{2022^2}{x}}+4044}=\dfrac{1}{2..\sqrt{\dfrac{x.2022^2}{x}}+4044}=\dfrac{1}{4044+4044}=\dfrac{1}{8088}\)-\(A_{max}=\dfrac{1}{8088}\Leftrightarrow x=2022\)

1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)

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

2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)

\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)

4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) 

\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)

\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)

e: Ta có: \(E=\left(x-8\right)^2+\left(x+7\right)^2\)

\(=x^2-16x+64+x^2+14x+49\)

\(=2x^2-2x+113\)

\(=2\left(x^2-x+\dfrac{113}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{225}{4}\right)\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

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

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 \("="\Leftrightarrow x=-5\)