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

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

\(2x+\left|2x-5\right|=2x+\left|5-2x\right|\ge2x+5-2x=5.\Rightarrow A_{min}=5.\text{Dâu "=" xay }ra\Leftrightarrow2x-5\ge0\Leftrightarrow x\le2,5\)

\(M=\left|x\right|+\left|x-1\right|=\left|x\right|+\left|1-x\right|\ge x+1-x=1\Rightarrow M_{min}=1.\text{Dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le1\)

\(A=x-\sqrt{x}\Leftrightarrow A+\frac{1}{4}=x-\sqrt{x}+\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow A+\frac{1}{4}\ge0\Rightarrow A_{min}=\frac{-1}{4}.\text{Dâus "=" xay ra khi:}x=\frac{1}{4}\)

Bài 1:

Sửa đề :v

\(B=x\left(x-3\right)\left(x-1\right)\left(x-4\right)\)

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

Đặt \(x^2-4x=t\)

\(B=t\left(t+3\right)\)

\(B=t^2+3t=t^2+2\cdot t\cdot\frac{3}{2}+\frac{9}{4}-\frac{9}{4}=\left(t+\frac{3}{2}\right)^2-\frac{9}{4}\ge\frac{-9}{4}\forall t\)

Dấu "=" xảy ra \(\Leftrightarrow t=\frac{-3}{2}\Leftrightarrow x^2-4x=\frac{-3}{2}\Leftrightarrow x=\frac{4\pm\sqrt{10}}{2}\)

Bài 2: Mình nghĩ nên sửa đề tìm min \(A=\left|2x\right|+\left|2x-5\right|\)

Bài 3:

\(M=\left|x\right|+\left|x-1\right|\)

\(M=\left|x\right|+\left|1-x\right|\ge\left|x+1-x\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow x\left(1-x\right)\ge0\Leftrightarrow0\le x\le1\)

Bài 4:

\(A=x-\sqrt{x}\)

Do điều kiện \(x\ge0\)

\(\Rightarrow A\ge0+0=0\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

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

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

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

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