1/a+1/b+1/c=0

=>(ab+ac+bc)/abc=0

=> ab+ac+bc=0

(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=0

=> a^2+b^2+c^2=0

Bạn xem lại đề nhé.

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

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

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

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

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

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

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

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

\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b) Ta có:

\(A\cdot\sqrt{x}=25\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}=25\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=25\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=5^2\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=16\\\sqrt{x}=-6\text{(vô lý)}\end{matrix}\right.\) 

c) Ta xét hiệu:

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

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

\(A-4=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)

\(A-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)

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

Với \(x>0\) thì \(\left(\sqrt{x}-1\right)>0\) và \(\sqrt{x}>0\)

\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

Nên A > 4 (đpcm)

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

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

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

2: A*căn x=25

=>(căn x+1)^2=25

=>căn x+1=5

=>x=16

3: \(A-4=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

=>A>4

Với mọi số thực ta luôn có:

`(a-b)^2>=0`

`<=>a^2-2ab+b^2>=0`

`<=>a^2+b^2>=2ab`

`<=>2(a^2+b^2)>=(a+b)^2=1`

`<=>a^2+b^2>=1/2(đpcm)`

Dấu "=' `<=>a=b=1/2`

ta có:

(a²+b²)(1²+1²)≥(a.1+b.1)²

⇔ 2(a²+b²) ≥ (a+b)²

⇔ 2(a²+b²)≥ 1 (vì a+b=1)

⇔ a² +b² ≥ 1/2 (đpcm)

dấu "=) xảy ra khi a = b = 1/2

Có \(\dfrac{a}{b}=\dfrac{c}{d}=>ad=bc\) => a² = ad => a=d

Xét \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

<=> (a+b)(c-a) = (a-b)(c+a)

<=> (a+b)(c-d) = (a-b)(c+d)

<=> ac - ad + bc - bd = ac + ad -bc -bd

<=> 2bc = 2ad (luôn đúng) => đpcm

Bài 1:

Từ \(\dfrac{x}{2}=\dfrac{y}{7}=\dfrac{2x}{4}=\dfrac{5y}{35}\) và 2x-5y=93

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{7}=\dfrac{2x}{4}=\dfrac{5y}{35}=\dfrac{2x-5y}{4-35}=\dfrac{93}{-31}=-3\)

=> x = 2 * (-3) = -6

y = 7 * (-3) = -21

bài 2 tui ko hỉu

B1. Tìm x, y biết :

\(\dfrac{x}{2}\)=\(\dfrac{y}{7}\) và 2x - 5y = 93

⇒ \(\dfrac{2x}{2}\)=\(\dfrac{5y}{7}\)

Áp dụng tính chất của dãy tỉ số bằng nhau :

\(\dfrac{2x}{2}\)=\(\dfrac{5y}{7}\)=\(\dfrac{2x-5y}{2-7}\)=\(\dfrac{93}{-5}\)

Suy ra :

\(\dfrac{2x}{2}\)=\(\dfrac{93}{-5}\) ⇒ 2x = \(\dfrac{93}{-5}\). 2 = \(\dfrac{186}{-5}\)

⇒ x = \(\dfrac{186}{-5}\): 2

⇒ x = \(\dfrac{93}{-5}\)

⇒ x = -18.6

\(\dfrac{5y}{7}\)=\(\dfrac{93}{-5}\) ⇒ 5y = \(\dfrac{93}{-5}\). 7 = \(\dfrac{651}{-5}\)

⇒ y = \(\dfrac{651}{-5}\): 5

⇒ y = \(\dfrac{651}{-25}\)

⇒ y = -26.04

Vậy : x = -18.6, y = -26.04

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

Thấy tử và mẫu của phân số đều lớn hơn 0 => \(\dfrac{a^2+a+1}{a^2-a+1}>0\)

b)\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2a+1\right)+\left(c^2-2a+1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (luôn đúng với mọi a,b,c)

Dấu = xra khi a=b=c=1

b)

\(a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) ( Luôn đúng)

Dấu "=" xảy ra khi a=b=c=1

Có: \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)

\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\) (do \(\left(a+b+c\right)^2=a^2+b^2+c^2=1\))