(\(\dfrac{2}{3}\))³ - 4.(-1\(\dfrac{3}{4}\))² + (- \(\dfrac{2}{3}\))³

= (\(\dfrac{2}{3}\))³ + 4.( \(\dfrac{7}{4}\))² - (\(\dfrac{2}{3}\))³

= [ (\(\dfrac{2}{3}\))³ - (\(\dfrac{2}{3}\))³ ] - \(\dfrac{49}{4}\)

=- \(\dfrac{49}{4}\)

- 49/4

4072299/4048

cho mik câu trả lời cụ thể đc k bn

Do Oz là tia phân giác của góc xOy nên:

Oz sẽ cắt xOy thành hai góc bằng nhau 

\(\Rightarrow\widehat{yOz}=\widehat{zOx}=\dfrac{\widehat{xOy}}{2}=\dfrac{30^o}{2}=15^o\)

Vậy: \(\widehat{yOz}=15^o\)

minh tag dung cho

BĐT: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\Rightarrow m=\left|x-1\right|+\left|x-5\right|\)

\(=\left|x-1\right|+\left|-\left(x-5\right)\right|\)

\(=\left|x-1\right|+\left|5-x\right|\)

Theo BĐT ta có: \(m=\left|x-1\right|+\left|5-x\right|\ge\left|x-1+5-x\right|=4\)

Vậy: \(m_{min}=4\)

giá chiếc pizza rau củ sau khi đã giảm là:

139000x(100%-10%)=125100 (đồng)

giá chiếc pizza thập cẩm sau khi đã giảm là:

289000x(100%-10%)=260100 (đồng)

giá tiền bác lan phải trả là:

(125100+260100)x(100%-5%)=365940 (đồng)

bác lan được trả lại số tiền là:

500000-365940=134060(đồng)

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:

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

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

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

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

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

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