Bài 1: Lũy thừa

bạn nhập pt vào máy tính rồi nhấn shift slove = ,sẽ ra nghiệm là 0,5 .lấy 0,5 thể vào căn thức rồi nhân liên hợp là ok

a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)

b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)

c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)

                              \(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)

d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)

Ta xét 3 trường hợp :

* Nếu \(x>4\) thì \(x-3>1\Rightarrow\left(x-3\right)^{2010}>1\Rightarrow\left(x-3\right)^{2010}+\left(x+4\right)^{2012}>1\) mâu thuẫn.

* Nếu \(x< 3\) thì \(x-4< -1\Rightarrow\left(x-4\right)^{2010}>1\Rightarrow\left(x-3\right)^{2010}+\left(x+4\right)^{2012}>1\) mâu thuẫn.

* Nếu \(3< x< 4\) thì \(x-3>1\Rightarrow\left|x-3\right|,\left|x-4\right|\le1\Rightarrow\left(x-3\right)^{2010}< \left(x-3\right),\left(x-4\right)^{2012}\le\left(4-x\right)\) 

Do đó \(\left(x-3\right)^{2010}+\left(x-4\right)^{2012}< \left(x-3\right)+\left(4-x\right)=1\) cũng mâu thuẫn

Mặt khác, với \(x=3;x=4\) thì đẳng thức đúng. Vậy ta có điều phải chứng minh

tìm x à

Đúng rồi ạ, nhờ chị giúp dùm

\(M=\sqrt{\left(\frac{1}{25}\right)^{\left(-\frac{3}{2}\right)}-\left(\frac{1}{8}\right)^{\left(-\frac{2}{3}\right)}}=\sqrt{\left(5^{-2}\right)^{-\frac{3}{2}}-\left(2^{-3}\right)^{-\frac{2}{3}}}=\sqrt{5^3-2^2}=\sqrt{121}=11\)

\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\left\{\left[\left(\frac{b}{a}\right)^{-1}\left(\frac{b}{a}\right)^{\frac{1}{5}}\right]^{\frac{1}{7}}\right\}^{\frac{35}{4}}=\left[\left(\frac{b}{a}\right)^{-\frac{4}{5}}\right]=\frac{a}{b}\)

\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\sqrt[4]{\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{35}}=\sqrt[4]{\left(\frac{a}{b}\sqrt[5]{\frac{b}{a}}\right)^5}\)

\(=\sqrt[4]{\left(\frac{a}{b}\right)^5.\frac{b}{a}}=\sqrt[4]{\left(\frac{a}{b}\right)^4}=\frac{a}{b}\)

\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19\left(-3\right)^{-3}=\left(2^{-1}\right)^{-4}-\left(5^4\right)^{\frac{1}{4}}-\left[\left(\frac{3}{2}\right)^2\right]^{-\frac{3}{2}}+19.\frac{1}{\left(-3\right)^3}\)

                                                                                  \(=2^4-5-\left(\frac{3}{2}\right)^{-3}-\frac{19}{27}\)

                                                                                  \(=11-\left(\frac{2}{3}\right)^3-\frac{19}{27}=10\)

\(C=\left(0,5\right)^{-4}-625^{0,25}-\left(2\frac{1}{4}\right)^{-1\frac{1}{2}}+19.\left(-3\right)^{-3}\)

\(=\left(\frac{1}{2}\right)^{-4}-625^{\frac{1}{4}}-\left(\frac{9}{4}\right)^{-\frac{3}{2}}+19.\left(-3\right)^{-3}\)

\(=2^4-\sqrt[4]{625}-\left(\frac{4}{9}\right)^{\frac{3}{2}}+19.\left(\frac{1}{\left(-3\right)^3}\right)\)

=\(16-5-\sqrt[2]{\left(\frac{4}{9}\right)^3}+19.\frac{1}{-27}=11-\frac{8}{27}-\frac{19}{27}=10\)

đáp án đúng = 10

\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)

   \(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)

    \(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)

\(D=4^{3+\sqrt{2}}.2^{1-\sqrt{2}}.2^{-3-\sqrt{2}}=2^{6+2\sqrt{2}}.2^{-2-2\sqrt{2}}=2^4=16\)

\(F=4^{3+\sqrt{2}}.2^{1-\sqrt{2}}.2^{-3-\sqrt{2}}=2^{6+2\sqrt{2}}.2^{1-\sqrt{2}}.2^{-3-\sqrt{2}}\)

\(=2^{6+2\sqrt{2}+1-\sqrt{2}-3-\sqrt{2}}=2^4=16\)

\(F=\left(1-2\sqrt{\frac{a}{b}}+\frac{a}{b}\right):\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2=\left(1-\sqrt{\frac{a}{b}}\right)^2:\left(\sqrt{a}-\sqrt{b}\right)^2\)

                                                        \(=\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)

ĐK: \(ab\ge0;b\ne0\)

\(F=\left(1-2\sqrt{\frac{a}{b}}+\frac{a}{b}\right):\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2\)

\(=\left(\sqrt{\frac{a}{b}}-1\right)^2:\left(\sqrt{a}-\sqrt{b}\right)^2=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{b}.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{1}{b}\)