y^-1=1/y

\(A=1-\frac{1}{y}+\frac{1}{y^2}=\left[\left(\frac{1}{y}\right)^2-\frac{1}{y}+\frac{1}{4}\right]-\frac{1}{4}+1=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy \(MinA=\frac{3}{4}\)khi \(\frac{1}{y}-\frac{1}{2}=0\Leftrightarrow y=2\)

 Vẽ BE//AC ( E thuộc tia đối của tia CD ) => ABEC là hình bình hành => BE = AC = 12; CE = AB = 10. Hạ BH _I_ DE dễ thấy dt(BCE) = dt(ABD) ( vì có cùng đáy AB = CE, cùng chiều cao BH) => dt(ABCD) = dt(BDE) 
Đặt BH = x; DH = y; EH = z có: 
{ BH² + DH² = BD² 
{ BH² + EH² = BE² 
{ DH + EH = DE = CD + CE 
<=> 
{ x² + y² = 35² (1) 
{ x² + z² = 12² (2) 
{ y + z = 37 (3) 
(1) - (2) : y² - z ² = 35² - 12² = 1081 <=> (y + z)(y - z) = 1081 => y - z = 1081/(y + z) = 1081/37 (4) 
(3) + (4) : 2y = 37 + 1081/37 = 2450/37 => y = 1225/37 => y² = 1225²/37² 
Thay vào (1) : x² = 35² - 1225²/37² = (1295² - 1225²)/37² = 420²/37² => x = 420/37 
S(ABCD) = S(BDE) = BH.DE/2 = x(y + z)/2 = (420/37).(37/2) = 240 (đvdt)

k có hình ak bạn?

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

Và \(\sqrt{x+8+6\sqrt{x-1}}=\sqrt{x-1+6\sqrt{x-1}+9}=\left(\sqrt{x-1}-3\right)^2\)

Ok dễ nhé

ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\)

\(=\sqrt{4-2.2.\sqrt{x-1}+x-1}+\sqrt{x-1+2.\sqrt{x-1}.3+9}\)

\(=\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}\)\(=|2-\sqrt{x-1}|+|\sqrt{x-1}+3|\ge|2-\sqrt{x-1}+\sqrt{x-1}+3|=5\)

Dấu bằng xảy ra khi \(2-\sqrt{x-1}\ge0\Leftrightarrow\sqrt{x-1}\le2\Leftrightarrow x\le3\)

Vậy \(1\le x\le3\)

Nếu đúng cho nhé bạn.

\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

ĐK:\(x\ge3\)

\(pt\Leftrightarrow\sqrt{x^2-5x+6}-\sqrt{2}+\sqrt{x+1}-\sqrt{5}=\sqrt{x-2}-\sqrt{2}+\sqrt{x^2-2x-3}-\sqrt{5}\)

\(\Leftrightarrow\frac{x^2-5x+6-2}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x+1-5}{\sqrt{x+1}+\sqrt{5}}=\frac{x-2-2}{\sqrt{x-2}+\sqrt{2}}+\frac{x^2-2x-3-5}{\sqrt{x^2-2x-3}+\sqrt{5}}\)

\(\Leftrightarrow\frac{x^2-5x+4}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}=\frac{x-4}{\sqrt{x-2}+\sqrt{2}}+\frac{x^2-2x-8}{\sqrt{x^2-2x-3}+\sqrt{5}}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}-\frac{x-4}{\sqrt{x-2}+\sqrt{2}}-\frac{\left(x-4\right)\left(x+2\right)}{\left(x+2\right)\sqrt{x^2-2x-3}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{x-1}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{1}{\sqrt{x+1}+\sqrt{5}}-\frac{1}{\sqrt{x-2}+\sqrt{2}}-\frac{x+2}{\left(x+2\right)\sqrt{x^2-2x-3}+\sqrt{5}}\right)=0\)

Suy ra x-4=0 =>x=4

\(x-4\sqrt{x}+\)\(4+y-1+6\sqrt{y-1}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)

\(\Leftrightarrow x=4,y=10\)

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

\(\Leftrightarrow x-1+x-2=3\)

\(\Leftrightarrow2x-3=3\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=6:2=3\)

bạn nghia thiếu  trường hợp thiếu đkxđ nữa

\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

 = \

  = \

Cho  +  = \frac{1}{a+b} ;  . CMR
a)  
b)  +  = 

Cho  +  = \frac{1}{a+b} ;  . CMR

a)  

b)  +  =  

lưu ý chép kĩ nhé nguyenchieubao

 ai k cho mk thì mk cho lại