a)\(\log_{\frac{2}{x}}x^2-14\log_{16x}x^3+40\log_{4x}\sqrt{x}=0\)ĐKXĐ: x>0

\(\Leftrightarrow2\log_{\frac{2}{x}}x-42\log_{16x}+20\log_{4x}\sqrt{x}=0\)

\(\Leftrightarrow\frac{2}{\log_x\frac{2}{x}}-\frac{42}{\log_x16x}+\frac{20}{\log_x4x}=0\)

\(\Leftrightarrow\frac{2}{\log_x2-1}-\frac{42}{4\log_x2+1}+\frac{20}{2\log_x+1}=0\)

Đặt \(\log_x2=a\left(a\in R\right)\)

Thay vào pt:\(\frac{2}{a-1}-\frac{42}{4a+1}+\frac{20}{2a+1}=0\)

\(\Leftrightarrow2a^2-a+4=0\)(pt này vô nghiệm)

Vậy pt đã cho vô nghiệm

cái đó phải là \(-42\log_{16x}x\) nhé bạn

\(\log_{\frac{x}{2}}4x^2+2\log_{\frac{x^3}{8}}2x+\log_{2x}\frac{x^4}{4}=-\frac{14}{3}\)(ĐKXĐ:x>0)

\(\Leftrightarrow2\log_{\frac{x}{2}}2x+\frac{2}{3}\log_{\frac{x}{2}}2x+2\log_{2x}\frac{x^2}{2}=-\frac{14}{3}\)

\(\Leftrightarrow\frac{8}{3}\log_{\frac{x}{2}}2x+2\log_{2x}\frac{x^2}{2}=-\frac{14}{3}\)

Xét \(\log_{2x}\frac{x^2}{2}=\log_{2x}\frac{x^2}{4}\cdot2=2\log_{2x}\frac{x}{2}+\log_{2x}2=\frac{2}{\log_{\frac{x}{2}}2x}+\frac{1}{1+\log_2x}\)

Thay vào phương trình ta được:

\(\frac{8}{3}\log_{\frac{x}{2}}2x+2\left(\frac{2}{\log_{\frac{x}{2}}2x}+\frac{1}{1+\log_2x}\right)=-\frac{14}{3}\)

Đặt \(\log_2x=a\left(a\in R\right)\)

Xét

\(\log_{\frac{x}{2}}2x=\log_{\frac{x}{2}}2+\log_{\frac{x}{2}}x=\frac{1}{\log_2\frac{x}{2}}+\frac{1}{\log_x\frac{x}{2}}=\frac{1}{\log_2x-1}+\frac{1}{1-\log_x2}=\frac{1}{a-1}+\frac{1}{1-\frac{1}{a}}=\frac{a+1}{a-1}\)

Thay vào pt ta được:

\(\frac{8}{3}\cdot\frac{a+1}{a-1}+2\left(2\cdot\frac{a-1}{a+1}+\frac{1}{a+1}\right)=-\frac{14}{3}\)

Giải ra ta được a=0 hoặc a=-23/17

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2^{-\frac{23}{17}}\end{matrix}\right.\)

Điều kiện x>1

Từ (1) ta có  \(\log_{\sqrt{3}}\frac{x+1}{x-1}>\log_34\) \(\Leftrightarrow\frac{x+1}{x-1}>2\) \(\Leftrightarrow\) 1<x<3

Đặt \(t=\log_2\left(x^2-2x+5\right)\)

Tìm điều kiện của t :

- Xét hàm số \(f\left(x\right)=\log_2\left(x^2-2x+5\right)\) với mọi x thuộc (1;3)

- Đạo hàm : \(f\left(x\right)=\frac{2x-2}{\ln2\left(x^2-2x+5\right)}>\) mọi \(x\in\left(1,3\right)\)

Hàm số đồng biến nên ta có \(f\left(1\right)\) <\(f\left(x\right)\) <\(f\left(3\right)\) \(\Leftrightarrow\)2<2<3

- Ta có \(x^2-2x+5=2'\)

 \(\Leftrightarrow\) \(\left(x-1\right)^2=2'-4\)

Suy ra ứng với mõi giá trị \(t\in\left(2,3\right)\) ta luôn có 1 giá trị \(x\in\left(1,3\right)\)

Lúc đó (2) suy ra : \(t-\frac{m}{t}=5\Leftrightarrow t^2-5t=m\)

Xét hàm số : \(f\left(t\right)=t^2-5t\) với mọi \(t\in\left(2,3\right)\)

- Đạo hàm : \(f'\left(t\right)=2t-5=0\Leftrightarrow t=\frac{5}{2}\)

- Bảng biến thiên :

Để hệ có 2 cặp nghiệm phân biệt \(\Leftrightarrow-6>-m>-\frac{25}{4}\)\(\Leftrightarrow\)\(\frac{25}{4}\) <m<6

\(\dfrac{1}{5}\)

-1

1.

\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)

\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)

\(=\dfrac{27}{4}log_2a\)

2.

\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)

\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)

a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)

b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)

c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)

d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)

                   \(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)

                   \(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)

\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)

\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)

\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)

\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)

\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)

\(\Leftrightarrow4x+y\le5\)

\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)

\(P_{max}=1\) khi \(x=y=1\)