a) \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)

\(\Rightarrow\left(\dfrac{1}{3}\right)^m=\left(\dfrac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

b) \(\dfrac{1}{9}\cdot27^n=3^n\)

\(\Rightarrow\dfrac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)

\(\Rightarrow\dfrac{3^{3n}}{3^2}=3^n\)

\(\Rightarrow3^{3n-2}=3^n\)

\(\Rightarrow3n-2=n\)

\(\Rightarrow2n=2\)

\(\Rightarrow n=1\)

c) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow\dfrac{2^3}{2^n}=2\)

\(\Rightarrow2^{3-n}=2^1\)

\(\Rightarrow3-n=1\)

\(\Rightarrow n=2\)

d) \(32^n\cdot16^{-n}=1024\)

\(\Rightarrow\left(2^5\right)^n\cdot\left(2^4\right)^{-n}=2^{10}\)

\(\Rightarrow2^{5n-4n}=2^{10}\)

\(\Rightarrow2^n=2^{10}\)

\(\Rightarrow n=10\)

e) \(3^{-1}\cdot3^n+5\cdot3^{n-1}=162\)

\(\Rightarrow3^{n-1}+5\cdot3^{n-1}=162\)

\(\Rightarrow3^{n-1}\cdot6=162\)

\(\Rightarrow3^{n-1}=27\)

\(\Rightarrow3^{n-1}=3^3\)

\(\Rightarrow n-1=3\)

\(n=4\)

f) \(\left(n-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(n-\dfrac{2}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow n-\dfrac{2}{3}=\dfrac{1}{3}\)

\(\Rightarrow n=\dfrac{1}{3}+\dfrac{2}{3}\)

\(\Rightarrow n=1\)

1

Lời giải:

$a^2+ab=c^2+bc$

$\Rightarrow a(a+b)=c(b+c)\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}(1)$

$a^2+ac=b^2+bc$

$\Rightarrow a(a+c)=b(b+c)\Rightarrow \frac{a+c}{b}=\frac{b+c}{a}(2)$
Từ $(1); (2)\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}+\frac{c+a}{b}$

Áp dụng TCDTSBN:

$\frac{a+b}{c}=\frac{b+c}{a}+\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2(a+b+c)}{a+b+c}=2$
$\Rightarrow a+b=2c; b+c=2a; c+a=2b$

$\Rightarrow a+b-(b+c)=2c-2a$

$\Rightarrow a-c=2c-2a\Rightarrow 3a=3c\Rightarrow a=c$
$2b=c+a=a+a=2a\Rightarrow a=b$

Vậy $a=b=c$

Do đó:

$K=(1+\frac{a}{a})(1+\frac{a}{a})(1+\frac{a}{a})=(1+1)(1+1)(1+1)=8$

\(M=\dfrac{\left(8a-3b\right)\left(2a+b\right)-\left(2a-b\right)\left(2a-5b\right)}{4a^2-b^2}=\)

\(=\dfrac{16a^2+2ab-3b^2-4a^2+12ab-5b^2}{4a^2-b^2}=\)

\(=\dfrac{12a^2+14ab-8b^2}{4a^2-b^2}=\)

\(=\dfrac{4a^2+14ab-6b^2+8a^2-2b^2}{4a^2-b^2}=\)

\(=\dfrac{2\left(2a^2+7ab-3b^2\right)+2\left(4a^2-b^2\right)}{\left(4a^2-b^2\right)}=2\)

a/

Ta có

\(DN\perp HA\left(gt\right);BC\perp HA\left(gt\right)\) => DN//BC

\(\Rightarrow\widehat{NDB}+\widehat{CBD}=180^o\) (Hai góc trong cùng phía bù nhau)

\(\Rightarrow\widehat{NDA}+\widehat{ADB}+\widehat{ABD}+\widehat{ABC}=180^o\)

Ta có

tg ABD vuông cân tại A \(\Rightarrow\widehat{ADB}=\widehat{ABD}=45^o\Rightarrow\widehat{ADB}+\widehat{ABD}=90^o\)

\(\Rightarrow\widehat{NDA}+\widehat{ABC}=180^o-90^o=90^o\)

Xét tg vuông ABH

\(\widehat{BAH}+\widehat{ABC}=90^o\)

\(\Rightarrow\widehat{NDA}=\widehat{BAH}\)

Xét tg vuông NDA và tg vuông BAH có

\(\widehat{NDA}=\widehat{BAH}\left(cmt\right)\)

AD=AB (cạnh bên tg cân)

=> tg NDA = tg BAH (Hai tg vuông có cạnh huyền và góc nhọn tương ứng bằng nhau)

=> DN = AH

C/m tương tự ta cũng có tg vuông MAE = tg vuông CHA => EM=AH

b/

Ta có

\(DN\perp HA\left(gt\right);EM\perp HA\left(gt\right)\) => DN//EM

Xét tg vuông DIN và tg vuông EIM có

DN=EM (cùng bằng AH)

\(\widehat{IDN}=\widehat{IEM}\) (góc so le trong)

=> tg DIN = tg EIM (Hai tg vuông có 1 cạnh góc vuông và góc nhọn tương ứng bằng nhau)

=> DI=IE

\(A=-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}\)

\(\Rightarrow-A=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{10-9}{9.10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\Rightarrow A=-\dfrac{2}{5}\)

Ta có : \(\dfrac{x}{3}=\dfrac{y}{-2}\Rightarrow\dfrac{3x}{9}=\dfrac{2y}{-4}\)  và `3x-2y=26`

ADTC dãy tỉ số bằng nhau ta có : 

\(\dfrac{3x}{9}=\dfrac{2y}{-4}=\dfrac{3x-2y}{9-\left(-4\right)}=\dfrac{26}{13}=2\\ \Rightarrow\dfrac{x}{3}=2\Rightarrow x=2\cdot3=6\\ \Rightarrow\dfrac{y}{-2}=2\Rightarrow y=2\cdot\left(-2\right)=-4\)

Ta có : $\genfrac{}{}{0ex}{}{x}{3}=\genfrac{}{}{0ex}{}{y}{-2}\Rightarrow \genfrac{}{}{0ex}{}{3x}{9}=\genfrac{}{}{0ex}{}{2y}{-4}$  và `3x-2y=26`

ADTC dãy tỉ số bằng nhau ta có : 

$\genfrac{}{}{0ex}{}{3x}{9}=\genfrac{}{}{0ex}{}{2y}{-4}=\genfrac{}{}{0ex}{}{3x-2y}{9-\left(-4\right)}=\genfrac{}{}{0ex}{}{26}{13}=2\Rightarrow \genfrac{}{}{0ex}{}{x}{3}=2\Rightarrow x=2\cdot 3=6\Rightarrow \genfrac{}{}{0ex}{}{y}{-2}=2\Rightarrow y=2\cdot \left(-2\right)=-4$

Lời giải:
Đặt $\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=a$

$\Rightarrow x=2a+1; y=4a-3; z=6a+5$

Thay vào điều kiện $5z-3x-4y=50$ thì:

$5(6a+5)-3(2a+1)-4(4a-3)=50$

$\Rightarrow 8a-16=0$

$\Rightarrow a=2$

Do đó:

$x=2a+1=2.2+1=5$

$y=4a-3=4.2-3=5$

$z=6a+5=6.2+5=17$

Đề bài là gì thế bạn?