ta có : 1/2<6/x<3/4

       Hay 6/12<6/x<6/8

=> xϵ{11;10;9}

A

Cần đáp số thoi

a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

Chọn A

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6

Sửa đề: Tìm các giá trị nguyên X thỏa mãn

3: \(\frac{-5}{11}<\frac{9}{x}<\frac{-5}{12}\)

=>\(\frac{-45}{99}<\frac{-45}{-5x}<\frac{-45}{108}\)

=>\(\frac{45}{99}>\frac{45}{-5x}>\frac{45}{108}\)

=>99<-5x<108

=>\(-\frac{99}{5}>x>-\frac{108}{5}\)

mà x nguyên

nên x∈{-20;-21}

4: Ta có: \(\frac{-11}{13}<\frac{9}{x}<\frac{-11}{15}\)

=>\(\frac{-99}{117}<\frac{-99}{-11x}<\frac{-99}{135}\)

=>\(\frac{99}{117}>\frac{99}{-11x}>\frac{99}{135}\)

=>117<-11x<135

=>\(-\frac{117}{11}>x>-\frac{135}{11}\)

mà x nguyên

nên x∈{-11;-12}

5: Ta có: \(\frac{-4}{5}<\frac{9}{x}<\frac{-4}{7}\)

=>\(\frac{-36}{45}<\frac{-36}{-4x}<\frac{-36}{28}\)

=>\(\frac{36}{45}>\frac{36}{-4x}>\frac{36}{28}\)

=>45<-4x<28

=>\(-\frac{45}{4}>x>-\frac{28}{4}\)

mà x nguyên

nên x∈∅

Đặt \(\left(\dfrac{x}{6};\dfrac{y}{3};\dfrac{z}{2}\right)=\left(a;b;c\right)\Rightarrow2^{6a}+4^{3b}+8^{2c}=4\)

\(\Leftrightarrow64^a+64^b+64^c=4\)

Áp dụng BĐT Cô-si:

\(4=64^a+64^b+64^c\ge3\sqrt[3]{64^{a+b+c}}\Rightarrow64^{a+b+c}\le\dfrac{64}{27}\)

\(\Rightarrow a+b+c\le log_{64}\left(\dfrac{64}{27}\right)\Rightarrow M=log_{64}\left(\dfrac{64}{27}\right)\)

Lại có: \(x;y;z\ge0\Rightarrow a;b;c\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}64^a\ge1\\64^b\ge1\\64^c\ge1\end{matrix}\right.\) \(\Rightarrow\left(64^b-1\right)\left(64^c-1\right)\ge0\)

\(\Rightarrow64^{b+c}+1\ge64^b+64^c\) (1)

Lại có: \(b+c\ge0\Rightarrow64^{b+c}\ge1\Rightarrow\left(64^a-1\right)\left(64^{b+c}-1\right)\ge0\)

\(\Rightarrow64^{a+b+c}+1\ge64^a+64^{b+c}\) (2)

Cộng vế (1);(2) \(\Rightarrow4=64^a+64^b+64^c\le64^{a+b+c}+2\)

\(\Rightarrow64^{a+b+c}\ge2\Rightarrow a+b+c\ge log_{64}2\)

\(\Rightarrow N=log_{64}2\)

\(\Rightarrow T=2log_{64}\left(\dfrac{64}{27}\right)+6log_{64}\left(2\right)\approx1,4\)

 thầy ơi giải như này được ko ạ?