Toán Lớp 9: tim x,y,z biet rang x+y+z+4=2√(x-2)+4 √(y-3) +6 √(z-5)

Question

Toán Lớp 9:  tim x,y,z biet rang  x+y+z+4=2√(x-2)+4 √(y-3) +6 √(z-5)

truongankimtrong · Answer

Điều kiện x&aacute;c định $x ge 2$, $y ge 3$, $z ge 5$   $ begin{array}{l} x + y + z + 4 = 2 sqrt {x - 2}  + 4 sqrt {y - 3}  + 6 sqrt {z - 5}      Leftrightarrow x - 2 sqrt {x - 2}  + y - 4 sqrt {y - 3}  + z - 6 sqrt {z - 5}  = 0     Leftrightarrow  left( {x - 2 - 2 sqrt {x - 2}  + 1}  right) +  left( {y - 3 - 4 sqrt {y - 3}  + 4}  right) +  left( {z - 5 - 6 sqrt {z - 5}  + 9}  right) = 0     Leftrightarrow { left( { sqrt {x - 2}  - 1}  right)^2} + { left( { sqrt {y - 3}  - 2}  right)^2} + { left( { sqrt {z - 5}  - 3}  right)^2} = 0     Leftrightarrow  left {  begin{array}{l}  sqrt {x - 2}  = 1    sqrt {y - 3}  = 2    sqrt {z - 5}  = 3  end{array}  right.  Leftrightarrow  left {  begin{array}{l} x - 2 = 1   y - 3 = 4   z - 5 = 9  end{array}  right.  Leftrightarrow  left {  begin{array}{l} x = 3   y = 7   z = 14  end{array}  right.( text{thỏa m&atilde;n})  end{array}$

thucquyenlan · Answer

Lời giải và giải thích chi tiết: ĐKXĐ : x>= 2 ; y >= 3 ; z >= 5 x+y+z+4=2sqrt(x-2)+4sqrt(y-3)+6sqrt(z-5) <=> x-2sqrt(x-2)+y-4sqrt(y-3)+z-6(sqrtz-5)+4=0 <=> (x-2-2sqrt(x-2)+1)+(y-3-4sqrt(y-3)+4)+(z-5-6sqrt(z-5)+9)=0 <=> (sqrt(x-2)-1)^2+(sqrt(y-3)+2)^2+(sqrt(z-5)-3)^2=0 Vì : {((sqrt(x-2)-1)^2 ge0 AA x),((sqrt(y-3)-2)^2 ge0 AAy),((sqrt(z-5)-3)^2 ge 0 AAz):} to (sqrt(x-2)-1)^2+(sqrt(y-3)-2)^2+(sqrt(z-5)-3)^2 ge 0 Mà : (sqrt(x-2)-1)^2+(sqrt(y-3)-2)^2+(sqrt(z-5)-3)^2=0 to Dấu '=' xảy ra : {(sqrt(x-2)-1=0),(sqrt(y-3)-2=0),(sqrt(z-5)-3=0):} <=> {(sqrt(x-2)=1),(sqrt(y-2)=2),(sqrt(z-5)=3):} <=> {(x-2=1),(y-3=4),(z-5=9):} <=> {(x=3),(y=7),(z=14):} text{(tmđk)} Vậy (x;y;z)=(3;7;14)