Bài 1: Tính a,\sqrt(7-2\sqrt(10))+ \sqrt(2) b,(\sqrt(3+\sqrt(5))-\sqrt(3-\sqrt(5)))^(2) Bài 2: Phân tích đa thức thành nhân tử a,x\sqrt(x)+8 b, \sqrt(x)-x^(2) c, x -5\sqrt(x) + 6 Giúp mình với mn ơi

nhung2000

2 năm trước

Loga Sinh Học lớp 12

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Thích Trả lời

ThaoPhan

Đáp án:

$\begin{array}{l}
1)\\
a)\sqrt 5 \\
b)2\\
2)\\
a)\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)\\
b)\sqrt x \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x + x} \right)\\
c)\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
1)\\
a)\sqrt {7 - 2\sqrt {10} } + \sqrt 2 \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} + \sqrt 2 \\
= \left| {\sqrt 5 - \sqrt 2 } \right| + \sqrt 2 \\
= \sqrt 5 - \sqrt 2 + \sqrt 2 \\
= \sqrt 5 \\
b){\left( {\sqrt {3 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } } \right)^2}\\
= 3 + \sqrt 5 - 2\sqrt {3 + \sqrt 5 } .\sqrt {3 - \sqrt 5 } + 3 - \sqrt 5 \\
= 6 - 2\sqrt {\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)} \\
= 6 - 2\sqrt {{3^2} - {{\left( {\sqrt 5 } \right)}^2}} \\
= 6 - 2\sqrt {9 - 5} \\
= 6 - 2\sqrt 4 \\
= 2\\
2)\\
a)x\sqrt x + 8\\
= {\left( {\sqrt x } \right)^3} + {2^3}\\
= \left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)\\
b)\sqrt x - {x^2}\\
= \sqrt x \left( {1 - x\sqrt x } \right)\\
= \sqrt x \left( {1 - {{\left( {\sqrt x } \right)}^3}} \right)\\
= \sqrt x \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x + x} \right)\\
c)x - 5\sqrt x + 6\\
= {\left( {\sqrt x } \right)^2} - 2\sqrt x - 3\sqrt x + 6\\
= \left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)
\end{array}$

Vote (0) Phản hồi (0) 2 năm trước

akenngu82

Đáp án:

`a)sqrt{7-2sqrt{10}}+sqrt2`

`=sqrt{5-2sqrt2.sqrt5+2}+sqrt2`

`=sqrt{(sqrt5-sqrt2)^2}+sqrt2`

`=sqrt5-sqrt2+sqrt2`

`=sqrt5`

`b)(sqrt{3+sqrt5}-sqrt{3-sqrt5})^2`

`=(sqrt{(6+2sqrt5)/2}-sqrt{(6-2sqrt5)/2})^2`

`=(sqrt{(5+2sqrt5+1)/2}-sqrt{(5-2sqrt5+1)/2})^2`

`=(sqrt{(sqrt5+1)^2/2}-sqrt{(sqrt5-1)^2/2})^2`

`=((sqrt5+1)/sqrt2-(sqrt5-1)/sqrt2)^2`

`=((sqrt5+1-sqrt5+1)/sqrt2)^2`

`=(sqrt2)^2=2`

`2a)xsqrtx+8(x>=0)`

`=(sqrtx)^3+2^3`

`=(sqrtx+2)(x-2sqrtx+4)`

`2b)sqrtx-x^2(x>=0)`

`=sqrtx(1-xsqrtx)`

`=sqrtx(1-sqrtx)(x+sqrtx+1)`

`2c)x-5sqrtx+6`

`=x-3sqrtx-2sqrtx+6`

`=sqrtx(sqrtx-3)-2(sqrtx-3)`

`=(sqrtx-3)(sqrtx-2).`

Vote (0) Phản hồi (0) 2 năm trước

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