1911

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Bài 3:

a) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a+1}\right)}\right)\)

\(=\dfrac{a-1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{1}{\sqrt{a}-1}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)

\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)

\(=\dfrac{a-1}{\sqrt{a}}\)

b) Thay \(a=3+2\sqrt{2}\) vào biểu thức A:

Ta có: \(\dfrac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}=\dfrac{2+2\sqrt{2}}{\sqrt{\left(1+2\sqrt{2}\right)^2}}=\dfrac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=2\)

Vậy giá trị biểu thức A tại \(a=3+2\sqrt{2}\)

1) \(9^{x-1}=\dfrac{1}{9}\) (1)

\(\Leftrightarrow3^{2x-2}=3^{-2}\)

\(\Leftrightarrow2x-2=-2\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{0\right\}\)

2) \(\dfrac{1}{3}:\sqrt{7-3x^2}=\dfrac{2}{15}\) (2)

\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{1}{\sqrt{7-3x^2}}=\dfrac{2}{15}\)

\(\Leftrightarrow\dfrac{1}{3\sqrt{7-3x^2}}=\dfrac{2}{15}\)

\(\Leftrightarrow15=6\sqrt{7-3x^2}\)

\(\Leftrightarrow6\sqrt{7-3x^2}=15\)

\(\Leftrightarrow\sqrt{7-3x^2}=\dfrac{5}{2}\)

\(\Leftrightarrow7-3x^2=\dfrac{25}{4}\)

\(\Leftrightarrow-3x^2=\dfrac{25}{4}-7\)

\(\Leftrightarrow-3x^2=-\dfrac{3}{4}\)

\(\Leftrightarrow x^2=\dfrac{1}{4}\)

\(\Leftrightarrow x=\pm\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

Đề sai

2;5;8;11;14;17;20;23;26;29;32;35;38;41

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