Bài 2 : 

a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = 2 

b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)

Dấu ''='' xảy ra khi x = -1 

 Bài 1 : 

a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)

c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

`x^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2 +3/4`

Vì `(x+1/2)^2 >= 0` với mọi `x`

  `=>(x+1/2)^2 +3/4 >= 3/4` với mọi `x`

 `=>` Biểu thức Min `=3/4<=>x=-1/2`

_____________

`(x-3)(x+5)+4=x^2+2x-11=x^2+2x+1-12=(x+1)^2-12`

  Vì `(x+1)^2 >= 0` với mọi `x`

    `=>(x+1)^2-12 >= -12` với mọi `x`

 `=>` Biểu thức Min `=-1/2<=>x=-1`

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)

mik chưa học giá trị lớn nhất là max và giá trị nhỏ nhất là min nên bạn cho mik kí hiệu khác nha

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

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

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

d) Ta có: \(D=x^2-2x+2\)

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

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

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Ai giúp mik vs

Huhu ai giúp vs

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

1) \(\sqrt{4+x}=2-x\) (ĐK: \(x\ge-4\))

\(\Leftrightarrow4+x=\left(2-x\right)^2\)

\(\Leftrightarrow4+x=4-4x+x^2\)

\(\Leftrightarrow x^2-4x-x+4-4=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

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

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

Vậy: \(S=\left\{0;5\right\}\)

2) 

a) ĐKXĐ: \(a>0,a\ne1\)

\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)

\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\cdot\dfrac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}+1}\)

\(A=\sqrt{a}\left(\sqrt{a}-1\right)\)

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

b) Ta có:

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

\(A=\left(\sqrt{a}\right)^2-2\cdot\dfrac{1}{2}\cdot\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

Mà: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi:

\(\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=-\dfrac{1}{4}\)

\(\Leftrightarrow a=\dfrac{1}{4}\)

Vậy: \(A_{min}=-\dfrac{1}{4}\)khi \(a=\dfrac{1}{4}\)