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