Toán Tìm x,y,z biết: `x^2 – 2x + y^2+ 4y + 5+ (2z-3)^2 =0` 16/07/2021 By Natalia Tìm x,y,z biết: `x^2 – 2x + y^2+ 4y + 5+ (2z-3)^2 =0`
Đáp án + Giải thích các bước giải: `x^2-2x+y^2+4y+5+(2z-3)^2=0` `<=> x^2-2x+1+y^2+4y+4+(2z-3)^2=0` `<=> (x-1)^2+(y+2)^2+(2z-3)^2=0` TH1: `(x-1)^2=0` `<=> x=1` TH2: `(y+2)^2=0` `<=> y=-2` TH3: `(2z-3)^2=0` `<=> 2z-3=0` `<=> z=3/2` `KL:…` Trả lời
`x^2 – 2x + y^2 + 4y + 5 + (2z – 3)^2 = 0` `=> (x^2 – 2x + 1) + (y^2 + 4y + 4) + (2z – 3)^2 = 0` Ta có: ` x^2 – 2x + 1 = x^2 – x – x + 1 = x. (x – 1) – (x – 1) = (x – 1)^2` `y^2 + 4y + 4 = y^2 + 2y + 2y + 4 = y(y + 2) + 2(y + 2) = (y + 2)^2` `=> (x – 1)^2 + (y + 2)^2 + (2z – 3)^2 = 0` Vì \(\left\{\begin{matrix}(x – 1)^2 \geq 0\forall x\\(y+2)^2 \geq 0 \forall y\\(2z-3)^2\geq0\forall z\end{matrix}\right.\) `=> (x – 1)^2 + (y+2)^2 + (2z-3)^2 >= 0 \forall x, y, z` Dấu `”=”` xảy ra `<=>` \(\left\{\begin{matrix}x – 1=0\\y+2=0\\2z-3=0\end{matrix}\right.\) `<=>` \(\left\{\begin{matrix}x=1\\y=-2\\z=1,5\end{matrix}\right.\) Vậy `x = 1; y = -2; z = 1,5` Trả lời
Đáp án + Giải thích các bước giải:
`x^2-2x+y^2+4y+5+(2z-3)^2=0`
`<=> x^2-2x+1+y^2+4y+4+(2z-3)^2=0`
`<=> (x-1)^2+(y+2)^2+(2z-3)^2=0`
TH1: `(x-1)^2=0`
`<=> x=1`
TH2: `(y+2)^2=0`
`<=> y=-2`
TH3: `(2z-3)^2=0`
`<=> 2z-3=0`
`<=> z=3/2`
`KL:…`
`x^2 – 2x + y^2 + 4y + 5 + (2z – 3)^2 = 0`
`=> (x^2 – 2x + 1) + (y^2 + 4y + 4) + (2z – 3)^2 = 0`
Ta có: ` x^2 – 2x + 1 = x^2 – x – x + 1 = x. (x – 1) – (x – 1) = (x – 1)^2`
`y^2 + 4y + 4 = y^2 + 2y + 2y + 4 = y(y + 2) + 2(y + 2) = (y + 2)^2`
`=> (x – 1)^2 + (y + 2)^2 + (2z – 3)^2 = 0`
Vì \(\left\{\begin{matrix}(x – 1)^2 \geq 0\forall x\\(y+2)^2 \geq 0 \forall y\\(2z-3)^2\geq0\forall z\end{matrix}\right.\)
`=> (x – 1)^2 + (y+2)^2 + (2z-3)^2 >= 0 \forall x, y, z`
Dấu `”=”` xảy ra
`<=>` \(\left\{\begin{matrix}x – 1=0\\y+2=0\\2z-3=0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x=1\\y=-2\\z=1,5\end{matrix}\right.\)
Vậy `x = 1; y = -2; z = 1,5`