\(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}\)

\(=\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\)

\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

.............

\(a,\)\(\sqrt{\left(x-1\right)\left(x-3\right)}\)

\(đkxđ\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)

\(\orbr{\begin{cases}x-1\ge0;x-3\ge0\\x-1< 0;x-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}x\ge1;x\ge3\\x< 1;x< 3\end{cases}\Rightarrow}\orbr{\begin{cases}x\ge3\\x< 1\end{cases}}}\)

\(b,\)\(\sqrt{\frac{4}{x+3}}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}x+3\ne0\\x+3\ge0\end{cases}\Rightarrow x+3>0}\)\(\Rightarrow x>-3\)

\(\sqrt{x^2-3x+2}=\sqrt{2}\)

\(\Rightarrow x^2-3x+2=2\)

\(\Rightarrow x^2-3x=0\)

\(\Rightarrow x=0;x=3\)

\(\sqrt{\text{ }\text{x}^2-3\text{x}+2}=\sqrt{2}\Leftrightarrow\text{x}^2-3\text{x}+2=2\Leftrightarrow\text{x}^2-3\text{x}=0\Leftrightarrow\text{x}\left(\text{x}-3\right)=0\Leftrightarrow\orbr{\begin{cases}\text{x}=0\\\text{x}=3\end{cases}}\)

\(3\sqrt{x}=15\)

\(\Leftrightarrow\sqrt{x}=5\)

\(\Leftrightarrow x=5^2\)

\(\Leftrightarrow x=25\)

Vậy x = 25

\(3\sqrt{x}=15\)

\(\Leftrightarrow\sqrt{x}=5\)

\(\Leftrightarrow x=5^2\)

\(\Leftrightarrow x=25\)

\(k\)\(nha\)

1 + 1 =2 

học tốt 

1+1=2

ko đăng linh tinh trên diễn đàn

học tốt

a,b, tự làm nha

c, y= ax + b (d' )

d // d' \(\Leftrightarrow\)\(\hept{\begin{cases}a=1\\b\ne2\end{cases}}\)

\(\Rightarrow\)d' :  y=x + b

thay x= 2 vào P ta đc

y=4

\(\Rightarrow\)điểm (2,4)

mà d' cắt P tại điểm có hđ = 2

\(\Rightarrow\)đ (2;4) \(\in\)d'

thay x=2, y=4 vào d' ta đc

4 = 2 + b

b= 2 ( ko tm)

\(\Rightarrow\)d' : y=x

#mã mã#

PT 

<=> \(x^2+x-3+\left(2x+1\right)\left(3-\sqrt{x^2+x+6}\right)=0\)

<=> \(x^2+x-3+\left(2x+1\right).\frac{-x^2-x+3}{3+\sqrt{x^2+x+6}}=0\)

<=> \(\orbr{\begin{cases}x^2+x-3=0\left(1\right)\\1-\frac{2x+1}{3+\sqrt{x^2+x+6}}=0\left(2\right)\end{cases}}\)

Giải (2)

\(2x-2=\sqrt{x^2+x+6}\)

<=> \(3x^2-9x-2=0\)với \(x\ge1\)

=> \(x=\frac{9+\sqrt{105}}{6}\)

Giải (1)

=> \(x=\frac{-1\pm\sqrt{13}}{2}\)

Vậy \(S=\left\{\frac{-1\pm\sqrt{13}}{2},\frac{9+\sqrt{105}}{6}\right\}\)

ĐKXĐ \(x\ge3\)

Đặt \(\sqrt{2x+3}+\sqrt{x-3}=t\left(t\ge0\right)\)

=> \(t^2=3x+2\sqrt{2x^2-3x-9}\left(1\right)\)

Khi đó Pt 

<=> \(t=t^2-6\)

=> \(t^2-t-6=0\)

Mà \(t\ge0\)

=> \(t=3\)

(1) <=> \(9-3x=2\sqrt{2x^2-3x-9}\)

=> \(9-3x\ge0\Rightarrow x\le3\)

Kết hợp với ĐKXĐ

=> \(x=3\)

Vậy  x=3

ĐKXĐ \(2x^2-1\ge0\)

PT <=> \(10x^2+3x-6=2\left(3x+1\right)\sqrt{2x^2-1}\)

<=> \(\left(3x+1\right)\left(x+2-2\sqrt{2x^2-1}\right)+7x^2-4x-8=0\)

<=> \(\left(3x+1\right).\frac{\left(x+2\right)^2-4\left(2x^2-1\right)}{x+2+2\sqrt{2x^2-1}}+7x^2-4x-8=0\)

<=> \(\left(3x+1\right).\frac{-7x^2+4x+8}{x+2+2\sqrt{2x^2-1}}+7x^2-4x-8=0\)

<=> \(\orbr{\begin{cases}-7x^2+4x+8=0\left(1\right)\\3x+1=x+2+2\sqrt{2x^2-1}\left(2\right)\end{cases}}\)

Giải (2)

\(2x-1=2\sqrt{2x^2-1}\)

<=> \(4x^2+4x-5=0\)với \(x\ge\frac{1}{2}\)

=> \(x=\frac{-1+\sqrt{6}}{2}\)

GIải (1)

\(x=\frac{2\pm2\sqrt{15}}{7}\)thỏa mãn ĐKXĐ

Vậy \(S=\left\{\frac{2\pm2\sqrt{15}}{7},\frac{-1+\sqrt{6}}{2}\right\}\)

ĐKXĐ \(x\ge\frac{1}{2}\)

Đặt \(\sqrt{x^2+2x}=a,\sqrt{2x-1}=b\left(a,b\ge0\right)\)

=> \(3a^2-b^2=3x^2+4x+1\)

Khi đó PT <=> 

\(a+b=\sqrt{3a^2-b^2}\)

=> \(a^2+2ab+b^2=3a^2-b^2\)

=> \(a^2-ab-b^2=0\)

=> \(a=\frac{1+\sqrt{5}}{2}.b\)

=> \(x^2+2x=\frac{6+2\sqrt{5}}{4}.\left(2x-1\right)\)

=> \(x=\frac{1+\sqrt{5}}{2}\)thỏa mãn ĐKXĐ

Vậy \(x=\frac{1+\sqrt{5}}{2}\)