Các bạn giúp mình nhanh nhanh với=((

kieuthanh72

2 năm trước

Loga Sinh Học lớp 12

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kimviphunhan1995

Đáp án:

\(\begin{array}{l}
1)\left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\\
2)\left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
3)x = 2\\
4)\left[ \begin{array}{l}
x = 3\\
x = - 4
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\sqrt {{x^2} + 9} = 5\\
\to {x^2} + 9 = 25\\
\to {x^2} = 16\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\\
2)\sqrt {{{\left( {2x - 5} \right)}^2}} = 1\\
\to \left| {2x - 5} \right| = 1\\
\to \left[ \begin{array}{l}
2x - 5 = 1\\
2x - 5 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
3)DK:x \ge 2\\
\sqrt {{x^2} - x - 2} = \sqrt {x - 2} \\
\to {x^2} - x - 2 = x - 2\\
\to {x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = 2
\end{array} \right.\\
\to x = 2\\
4)DK:\left[ \begin{array}{l}
x = 3\\
x \le - 3
\end{array} \right.\\
\sqrt {{x^2} - 9} = \sqrt {3 - x} \\
\to {x^2} - 9 = 3 - x\\
\to {x^2} + x - 12 = 0\\
\to \left( {x + 4} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 4
\end{array} \right.
\end{array}\)

Vote (0) Phản hồi (0) 2 năm trước

thuhoai2132002

Đáp án+Giải thích các bước giải:

`1)sqrt{x^2+9}=5`

`<=>x^2+9=25`

`<=>x^2=16`

`<=>[(x=4),(x=-4):}`

`=>S={4;-4}.`

`2)sqrt{4x^2-20x+25}=1`

`<=>sqrt{(2x-5)^2}=1`

`<=>|2x-5|=1`

`<=>[(2x-5=-1),(2x-5=1):}`

`<=>[(2x=4),(2x=6):}`

`<=>[(x=2),(x=3):}`

`=>S={2;3}`

`3)sqrt{x^2-x-2}=x-2(x>=2)`

`<=>x^2-x-2=(x-2)^2=x^2-4x+4`

`<=>3x=6`

`<=>x=2(tm)`

`4)sqrt{x^2-9}=sqrt{3-x}`

ĐK:`{(x^2-9>=0),(3-x>=0):}<=>{(x^2>=9),(x<=3):}`

`<=>{([(x>=3),(x<=-3):}),(x<=3):}<=>[(x=3),(x<=-3):}`

`x=3=>TM`

`x<=-3pt<=>x^2-9=3-x<=>x^2+x-12=0<=>(x+4)(x-3)=0<=>x=-4`

`=>S={3;-4}.`

Vote (0) Phản hồi (0) 2 năm trước

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