Bài 1: Lũy thừa

A sai luôn; do ab<0 nên \(lna\) và \(lnb\) không tồn tại

Do \(\dfrac{3}{4}\) không nguyên nên trước hết \(2-a>0\Rightarrow a< 2\)

Sau đó \(\left(2-a\right)^{\dfrac{3}{4}}>\left(2-a\right)^2\) mà \(2>\dfrac{3}{4}\Rightarrow2-a< 1\Rightarrow a>1\)

\(\Rightarrow1< a< 2\)

\(log_a\left(ac^2\right)=log_c\left(b^3c\right)\Leftrightarrow1+2log_ac=1+3log_cb\)

\(\Leftrightarrow\left\{{}\begin{matrix}2log_ac-3log_cb=0\\2log_ac+log_cb=8\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}log_ac=3\\log_cb=2\end{matrix}\right.\)

\(\Rightarrow P=log_ab+log_ca+2log_cb=\dfrac{log_cb}{log_ca}+log_ca+2log_cb=log_ca.log_cb+log_ca+2log_cb\)

x=8

\(P=\dfrac{1}{\dfrac{1}{2}log_ax}+\dfrac{1}{\dfrac{1}{4}log_ax}+...+\dfrac{1}{\dfrac{1}{2n}log_ax}\)

\(=\dfrac{2+4+...+2n}{log_ax}=\dfrac{n\left(n+1\right)}{log_ax}\)

\(log_{ab}b=\dfrac{1}{log_bab}=\dfrac{1}{log_ba+1}=\dfrac{1}{\dfrac{1}{log_ab}+1}=\dfrac{log_ab}{log_ab+1}\)

Đặt \(log_ab=x\) (cho gọn biểu thức)

\(P=\left(x+\dfrac{1}{x}+2\right)\left(x-\dfrac{x}{x+1}\right).\dfrac{1}{x}-1\)

\(=\left(\dfrac{x^2+2x+1}{x}\right)\left(\dfrac{x}{x+1}\right).\dfrac{1}{x}-1\)

\(=\dfrac{x+1}{x}-1=\dfrac{1}{x}=log_ba\)

Pt đầu tương đương: \(\sqrt[3]{x^2}+2\sqrt[3]{y^2}+4\sqrt[3]{z^2}=7\)

Pt 2 tương đương:

\(\left(xy^2+z^4\right)^2-\left(xy^2-z^4\right)^2=4\)

\(\Leftrightarrow4xy^2z^4=4\)

\(\Leftrightarrow xy^2z^4=1\) (1)

Quay lại pt đầu, áp dụng AM-GM:

\(7=\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}+\sqrt[3]{z^2}+\sqrt[3]{z^2}+\sqrt[3]{z}\ge7\sqrt[7]{\sqrt[3]{x^2}.\sqrt[3]{y^4}.\sqrt[3]{z^8}}\)

\(\Leftrightarrow\sqrt[21]{x^2y^4z^8}\le1\)

\(\Leftrightarrow x^2y^4z^8\le1\)

\(\Rightarrow\left|xy^2z^4\right|\le1\Rightarrow xy^2z^4\le1\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x^2=y^2=z^2\\xy^2z^4=1\\x>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\pm1\\z=\pm1\end{matrix}\right.\)

Các bộ thỏa mãn là: \(\left(1;1;1\right);\left(1;1;-1\right);\left(1;-1;1\right);\left(1;-1;-1\right)\)

Ta có:

\(\sqrt[3]{7}< \sqrt[3]{8}=2\) và \(\sqrt{15}< \sqrt{16}=4\), suy ra \(\sqrt[3]{7}+\sqrt{15}< 6\).

\(\sqrt{10}>\sqrt{9}=3\) và \(\sqrt[3]{28}>\sqrt[3]{27}=3\), suy ra \(\sqrt{10}+\sqrt[3]{28}>6\).

Vậy \(\sqrt[3]{7}+\sqrt{15}< \sqrt{10}+\sqrt[3]{28}\).