bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

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

\(A=\left(x^2-2x+1\right)+9\)

\(A=\left(x-1\right)^2+9\)

Mà  \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra khi :

\(x-1=0\Leftrightarrow x=1\)

Vậy Min A = 9 khi x = 1

\(B=x^2-5x-7\)

\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)

\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow B\ge-\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)

\(C=3x^2+3x-5\)

\(3C=9x^2+9x-15\)

\(3C=\left(9x^2+9x+\frac{9}{4}\right)-\frac{69}{4}\)

\(3C=\left(3x+\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(3x+\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow3C\ge-\frac{69}{4}\)

\(\Leftrightarrow C\ge-\frac{23}{4}\)

Dấu "=" xảy ra khi :

\(3x+\frac{3}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy ...

A=((x-3)+(x+1))^2>=0

A=(x-2)^2>=0

Dấu bằng xảy ra khi

(x-2)^2=0

x-2=0

x=0+2

x=2

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau

2/

b) ( cái bài này chịu)

c) (x+1)^3-(x-1)^3-6(x-1)^2=-10

(x+1-x+1)\(\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)\(-6\left(x^2-2x+1\right)=-10\)

\(2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6x^2+12x-6=-10\)

\(2\left(3x^2+1\right)-6x^2+12x-6=0\)

\(6x^2+2-6x^2+12x-6=-10\)

\(12x=-10+4\)

\(12x=-6=>x=-\frac{1}{2}\)

d) (5x-1)^2-(5x-4)(5x+4)=7

\(25x^2-10x+1-25x^2+16=7\)

-10x = 7 - 17

-10x = -10

x= 1

Câu còn lại bn làm tương tự

3/

a) 

Ta có: 

(a+b+c)^2=3(ab+bc+ca)

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 3ab - 3bc - 3ac = 0

a^2 + b^2 + c^2  - ac - bc - ab = 0

2a^2 + 2b^2 + 2c^2  - 2ac - 2bc - 2ab = 0

(a²-2ab+b²⁾+(a²-2ac+c²) + (b²-2bc +c²) = 0

(a-b)^2 + (a-c)^2 + (b-c)^2 =0

=> a=b=c

a) Mình ko rõ 

b) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12=2\)

\(\Leftrightarrow x^3+3x-13-x^3-27=2\)

\(\Leftrightarrow3x-40=2\)

\(\Leftrightarrow3x=42\)

\(\Leftrightarrow x=14\)

\(x\)(\(x\) - 2)^? vậy em ha

1 

\(\left(x-2\right):2.3=6\)

\(\Leftrightarrow\left(x-2\right):2=2\)

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

\(\Leftrightarrow x=4+2=6\)

c) ta có

\(\left[\left(2x+1\right)+1\right]m:2=625\)

\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)

\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)

\(\Leftrightarrow\left(2x+1\right)^2=1250\)

...

2

\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)

\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)