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

Bài 1: (1/2x - 5)²⁰ + (y² - 1/4)¹⁰ < 0 (1)

Ta có: (1/2x - 5)²⁰ \(\ge\)0 \(\forall\)x

         (y² - 1/4)¹⁰ \(\ge\)0 \(\forall\)y

=> (1/2x - 5)²⁰ + (y² - 1/4)¹⁰ \(\ge\)0 \(\forall\)x;y

Theo (1) => ko có giá trị x;y t/m

Bài 2. (x - 7)^{x + 1} - (x - 7)^{x + 11} = 0

=> (x - 7)^{x + 1}.[1 - (x - 7)¹⁰] = 0

=> \(\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)

=> x = 7

hoặc : \(\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)

=> x = 7

hoặc : \(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)

Bài 3a) Ta có: (2x + 1/3)⁴ \(\ge\)0 \(\forall\)x

=> (2x +1/3)⁴ - 1 \(\ge\)-1 \(\forall\)x

=>  A \(\ge\)-1 \(\forall\)x

Dấu "=" xảy ra <=> 2x + 1/3 = 0 <=> 2x = -1/3 <=> x = -1/6

Vậy Min A = -1 tại x = -1/6

b) Ta có: -(4/9x - 2/5)⁶ \(\le\)0 \(\forall\)x

=> -(4/9x - 2/15)⁶ + 3 \(\le\)3 \(\forall\)x

=> B \(\le\)3 \(\forall\)x

Dấu "=" xảy ra <=> 4/9x - 2/15 = 0 <=> 4/9x = 2/15 <=> x = 3/10

vậy Max B = 3 tại x = 3/10

Đúng ko vậy bạn

b: \(B=\sqrt{x^2-8x+18}-1\)

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

(x-4)^2+2>=2

=>\(\sqrt{\left(x-4\right)^2+2}>=\sqrt{2}\)

=>B>=căn 2-1

Dấu = xảy ra khi x=4

a: \(D=3+\sqrt{2x^2-8x+33}\)

\(=3+\sqrt{2\left(x^2-4x+\dfrac{33}{2}\right)}\)

\(=\sqrt{2\left(x^2-4x+4\right)+25}+3\)

\(=\sqrt{2\left(x-2\right)^2+25}+3>=5+3=8\)

Dấu = xảy ra khi x=2

Cứu

Huhu........

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

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

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

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

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

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

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

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

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

Trả lời:

Bài 1: a,

\(A=\left|x-1\right|+3\)

Vì \(\left|x-1\right|\ge0\forall x\)

\(\Rightarrow\left|x-1\right|+3\ge3\forall x\)

Dấu = xảy ra khi x - 1 = 0 \(\Leftrightarrow x=1\)

Vậy GTNN của A = 3 khi x = 1

\(B=\left|x-7\right|-4\)

Vì \(\left|x-7\right|\ge0\forall x\)

  \(\Rightarrow\left|x-7\right|-4\ge-4\forall x\)

Dấu = xảy ra khi x - 7 = 0 \(\Leftrightarrow x=7\)

Vậy GTNN của B = -4 khi x = 7

b, \(C=-\left|x-3\right|+2\)

Vì \(\left|x-3\right|\ge0\forall x\)

\(\Rightarrow-\left|x-3\right|\le0\forall x\)

\(\Rightarrow-\left|x-3\right|+2\le2\forall x\)

Dấu = xảy ra khi x - 3 = 0 \(\Leftrightarrow x=3\)

Vậy GTLN của C = 2 khi x = 3

1. B = | x - 2018 | + | x - 2019 | + | x - 2020 |

= ( | x - 2018 | + | x - 2020 | ) + | x - 2019 | 

= ( | x - 2018 | + | 2020 - x | ) + | x - 2019 |

Vì \(\hept{\begin{cases}\left|x-2018\right|+\left|2020-x\right|\ge\left|x-2018+2020-x\right|=2\\\left|x-2019\right|\ge0\end{cases}}\)=> B ≥ 2 ∀ x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-2018\right)\left(2020-x\right)\ge0\\x-2019=0\end{cases}}\Rightarrow x=2019\)

Vậy MinB = 2 <=> x = 2019

2. ĐKXĐ : x ≥ 0

Ta có : \(\sqrt{x}+3\ge3\forall x\ge0\)

=> \(\frac{2019}{\sqrt{x}+3}\le673\forall x\ge0\). Dấu "=" xảy ra <=> x = 0 (tm)

Vậy MaxC = 673 <=> x = 0

Bài 1 : 

\(B=\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)

Ta có : \(\left|x-2018\right|\ge0\forall x;\left|x-2019\right|\ge0\forall x;\left|x-2020\right|\ge0\forall x\)

\(\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\ge0\)

Dấu ''='' xảy ra khi \(x=2018;x=2019;x=2020\)

Vậy GTNN B là 0 khi x = 2018 ; x = 2019 ; x = 2020 

Bài 1

(2x + 9)² > 0

3(2x + 9)² > 0

3(2x + 9)² - 1 > - 1

Vậy GTNN của biểu thức là - 1

Bài 2

(x - a)(x + a) = x² - 169

x² - a² = x² - 169

a² = 169

mà a < 0

nên a = - 13