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Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\sqrt {81} – \sqrt {80} .\sqrt {0,2} = \sqrt {81} – \sqrt {80.0,2} = \sqrt {81} – \sqrt {16} = \sqrt {{9^2}} – \sqrt {{4^2}} = 9 – 4 = 5\\
b,\\
\sqrt {{{\left( {2 – \sqrt 5 } \right)}^2}} – \dfrac{1}{2}\sqrt {20} = \left| {2 – \sqrt 5 } \right| – \dfrac{1}{2}\sqrt {{2^2}.5} = \left( {\sqrt 5 – 2} \right) – \dfrac{1}{2}.2.\sqrt 5 = \sqrt 5 – 2 – \sqrt 5 = – 2\\
2,\\
a,\,\,\,\,\,\,\,\,\,\,\sqrt { – x + 1} \\
DKXD:\,\,\,\, – x + 1 \ge 0 \Leftrightarrow – \left( {x – 1} \right) \ge 0 \Leftrightarrow x – 1 \le 0 \Leftrightarrow x \le 1\\
b,\,\,\,\,\,\,\,\sqrt {\dfrac{1}{{{x^2} – 2x + 1}}} \\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
\dfrac{1}{{{x^2} – 2x + 1}} \ge 0\\
{x^2} – 2x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{{{\left( {x – 1} \right)}^2}}} \ge 0\\
{\left( {x – 1} \right)^2} \ne 0
\end{array} \right. \Leftrightarrow x \ne 1\\
2,\\
1,\\
a,\\
ab + b\sqrt a + \sqrt a + 1\\
= \left( {ab + b\sqrt a } \right) + \left( {\sqrt a + 1} \right)\\
= b\sqrt a .\left( {\sqrt a + 1} \right) + \left( {\sqrt a + 1} \right)\\
= \left( {\sqrt a + 1} \right)\left( {b\sqrt a + 1} \right)\\
2,\\
\sqrt {9x + 9} + \sqrt {x + 1} = 20\,\,\,\,\,\,\,\,\left( {x \ge – 1} \right)\\
\Leftrightarrow \sqrt {9\left( {x + 1} \right)} + \sqrt {x + 1} = 20\\
\Leftrightarrow 3.\sqrt {x + 1} + \sqrt {x + 1} = 20\\
\Leftrightarrow 4\sqrt {x + 1} = 20\\
\Leftrightarrow \sqrt {x + 1} = 5\\
\Leftrightarrow x + 1 = 25\\
\Leftrightarrow x = 24
\end{array}\)
Đáp án:
Giải thích các bước giải:
Bài 1.1:
a)√81-√80.√0,2
=9-4√5.1/√5
=9-4=5
b)√(2-√5)² -1/2.√20
=√5 – 2 – 1/2.2√5
=√5 – 2 – √5 =-2
Bài1.2:
a)√(-x+1) có nghĩa khi -x+1≥0
⇔x≤ 1
Vậy…
b)√(1/x²-2x+1)
⇔ √(1/x²-2x+1)≥0 ⇔√(1/(x-1)²)≥0 ⇒ x≠1
và x²-2x+1≠0 và (x-1)²≠0
Bài 2.1:
a)ab+ b√a +√a +1 (a≥0)
=b√a(√a +1)+(√a+1)
=(b√a +1)(√a +1)
b)4a+1(a<0)
=-4√a² + 1 (a<0)
=1-(2√a)²=(1-2√a)(1+2√a)
Bài 2.2:
√(9x+9) +√(x+1) = 20 (ĐK:x≥-1)
⇔3√(x+1) +√(x+1) = 20
⇔√(x+1) . (3+1)=20
⇔√x+1 =5
⇔x+1=25⇔x=24(tmđk)
Vậy…