a: \(A=2xy^2\cdot\left(\dfrac{1}{2}x^2y^2x\right)\)

\(=2\cdot\dfrac{1}{2}\cdot xy^2\cdot x^3y^2=x^4y^4\)

b: Bậc là 8

c: \(A=x^4y^4\)

Hệ số là 1

Phần biến là \(x^4;y^4\)

d: Khi x=1 và y=-1 thì \(A=1^4\cdot\left(-1\right)^4=1\)

e: \(x^4>0\forall x\ne0;y^4>0\forall y\ne0\)

Do đó: \(x^4\cdot y^4>0\forall x,y\ne0\)

=>A luôn dương khi x,y đều khác 0

a) 

\(A=2xy^2\cdot\left(\dfrac{1}{2}x^2y^2x\right)\\ =\left(2\cdot\dfrac{1}{2}\right)\cdot\left(x\cdot x^2\cdot x\right)\cdot\left(y^2\cdot y^2\right)\\ =x^4y^4\)

b) Bậc: 4 + 4 = 8

c) Hệ số là: 1

Phần biến là: `x^4y^4` 

d) Thay x = 1 và  y = -1 vào A ta có:

\(A=1^4\cdot\left(-1\right)^4=1\cdot1=1\)

e) Ta có: \(\left\{{}\begin{matrix}x^4>0\forall x>0\\y^4>0\forall y>0\end{matrix}\right.=>A=x^4y^4>0\cdot0=0\forall x,y>0\)

=> A luôn nhận giá trị nguyên khi x,y khác 0

câu g là 1 và 1/7 ạ?

\(1,\)

\(a,\dfrac{-9}{51}.\dfrac{17}{6}\)

\(=-1.\dfrac{1}{2}\)

\(=-\dfrac{1}{2}\)

\(b,\dfrac{-25}{32}.\left(-0,2\right)\)

\(=\dfrac{-25}{32}.-\dfrac{2}{10}\)

\(=\dfrac{-5}{16}.-\dfrac{1}{2}\)

\(=\dfrac{5}{32}\)

\(c,-15,2.3,5=-53,2\)

\(d,\dfrac{-8}{15}.1\dfrac{1}{4}\)

\(=\dfrac{-8}{15}.\dfrac{5}{4}\)

\(=\dfrac{-2}{3}.\dfrac{1}{1}\)

\(=\dfrac{-2}{3}\)

\(e,1\dfrac{2}{5}.\dfrac{-3}{14}\)

\(=\dfrac{7}{5}.\dfrac{-3}{14}\)

\(=\dfrac{1}{5}.\dfrac{-3}{2}\)

\(=-\dfrac{3}{10}\)

\(g,1\dfrac{1}{17}.1\dfrac{1}{36}\)

\(=\dfrac{18}{17}.\dfrac{37}{36}\)

\(=\dfrac{1}{17}.\dfrac{37}{2}\)

\(=\dfrac{37}{34}\)

\(#T.T\)

\(\left(1+\dfrac{1}{2}-\dfrac{1}{4}\right)^2\times\left(2+\dfrac{3}{7}\right)\\ =\left(\dfrac{4}{4}+\dfrac{2}{4}-\dfrac{1}{4}\right)^2\times\left(\dfrac{14}{7}+\dfrac{3}{7}\right)\\ =\left(\dfrac{5}{4}\right)^2\times\dfrac{17}{7}\\ =\dfrac{25}{16}\times\dfrac{17}{7}=\dfrac{425}{112}\)

Bạn nhấn vào biểu tượng Σ để nhập công thức toán học bạn nha!

\(#BecauseI'maStrongGirl\)

a) OA//IC => \(\widehat{CKB}=\widehat{AOK}=50^o\) (đồng vị) 

OB//DE => \(\widehat{CID}=\widehat{CKB}=50^o\) (đồng vị) 

b) Mà: \(\widehat{CIE}+\widehat{CID}=180^o\) (kề bù)

=> \(\widehat{CIE}=180^o-\widehat{CID}\)

=> \(\widehat{CIE}=180^o-50^o=130^o\)

Ta có: \(x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=6 \)

\(\Leftrightarrow x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-6=0\\ \Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(y^2+\dfrac{1}{y^2}-2\right)+\left(z^2+\dfrac{1}{z^2}-2\right)=0\\ \Leftrightarrow\left(x^2-2\cdot x^2\cdot\dfrac{1}{x^2}+\dfrac{1}{x^2}\right)+\left(y^2-2\cdot y^2\cdot\dfrac{1}{y^2}+\dfrac{1}{y^2}\right)+\left(z^2-2\cdot z^2\cdot\dfrac{1}{z^2}+\dfrac{1}{z^2}\right)=0\\ \Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+\left(z-\dfrac{1}{z}\right)^2=0\)

Mà: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\\\left(y-\dfrac{1}{y}\right)^2\ge0\forall y\\\left(z-\dfrac{1}{z}\right)^2\ge0\forall z\end{matrix}\right.=>\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+\left(z-\dfrac{1}{z}\right)^2\ge0\forall x,y,z\) 

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\\z^2=1\end{matrix}\right.\)  

\(P=x^{2024}+y^{2024}+z^{2024}\\=\left(x^2\right)^{1012}+\left(y^2\right)^{1012}+\left(z^2\right)^{1012}\\ =1^{1012}+1^{1012}+1^{1012}=3\)

a: \(8^{24}=\left(2^3\right)^{24}=2^{72};16^{20}=\left(2^4\right)^{20}=2^{80}\)

mà 72<80

nên \(8^{24}< 16^{20}\)

b: \(\left(-\dfrac{1}{25}\right)^{37}=-\left(\dfrac{1}{5}\right)^{74}=-\dfrac{1}{5^{74}};\left(-\dfrac{1}{125}\right)^{23}=-\dfrac{1}{\left(5^3\right)^{23}}=-\dfrac{1}{5^{69}}\)

\(5^{74}>5^{69}\)

=>\(\dfrac{1}{5^{74}}< \dfrac{1}{5^{69}}\)

=>\(-\dfrac{1}{5^{74}}>-\dfrac{1}{5^{69}}\)

=>\(\left(-\dfrac{1}{25}\right)^{37}>\left(-\dfrac{1}{125}\right)^{23}\)

c: \(A=\dfrac{3}{7^3}+\dfrac{5}{7^4}=\dfrac{3\cdot7+5}{7^4}=\dfrac{26}{7^4}\)

\(B=\dfrac{5}{7^3}+\dfrac{3}{7^4}=\dfrac{5\cdot7+3}{7^4}=\dfrac{38}{7^4}\)

mà 26<38

nên A<B

d: \(10A=\dfrac{10^8+10}{10^8+1}=1+\dfrac{9}{10^8+1}\)

\(10B=\dfrac{10^9+10}{10^9+1}=1+\dfrac{9}{10^9+1}\)

Ta có: \(10^8+1< 10^9+1\)

=>\(\dfrac{9}{10^8+1}>\dfrac{9}{10^9+1}\)

=>\(\dfrac{9}{10^8+1}+1>\dfrac{9}{10^9+1}+1\)

=>10A>10B

=>A>B

a: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{2}\right)^{10}\)

=>\(\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{10}\)

=>4x=10

=>x=2,5

b: \(\left(-\dfrac{8}{13}\right)^x=\dfrac{64}{169}\)

=>\(\left(-\dfrac{8}{13}\right)^x=\left(-\dfrac{8}{13}\right)^2\)

=>x=2

c: \(\left(\dfrac{1}{64}\right)^x=\left(-\dfrac{1}{8}\right)^{14}\)

=>\(\left(\dfrac{1}{64}\right)^x=\left(\dfrac{1}{64}\right)^7\)

=>x=7

d: \(\dfrac{27-x}{23}+\dfrac{28-x}{24}=\dfrac{29-x}{25}+\dfrac{30-x}{26}\)

=>\(\left(\dfrac{27-x}{23}-1\right)+\left(\dfrac{28-x}{24}-1\right)=\left(\dfrac{29-x}{25}-1\right)+\left(\dfrac{30-x}{26}-1\right)\)

=>\(\dfrac{4-x}{23}+\dfrac{4-x}{24}=\dfrac{4-x}{25}+\dfrac{4-x}{26}\)

=>\(\left(4-x\right)\left(\dfrac{1}{23}+\dfrac{1}{24}-\dfrac{1}{25}-\dfrac{1}{26}\right)=0\)

=>4-x=0

=>x=4

a: \(64^x:16^x=256\)

=>\(\left(\dfrac{64}{16}\right)^x=256\)

=>\(4^x=256=4^4\)

=>x=4

b: \(-\dfrac{2401}{7^x}=-7\)

=>\(\dfrac{2401}{7^x}=7\)

=>\(7^x=\dfrac{2401}{7}=343=7^3\)

=>x=3

c: \(\dfrac{625}{\left(-5\right)^x}=25\)

=>\(\left(-5\right)^x=\dfrac{625}{25}=25=\left(-5\right)^2\)

=>x=2

a) $64^x:16^x=256$

$\Rightarrow (4^3)^x:(4^2)^x=256$

$\Rightarrow (4^3:4^2)^x=256$

$\Rightarrow 4^x=4^4$

$\Rightarrow x=4$ (tmdk)

b) $\frac{-2401}{7^x}=-7$

$\Rightarrow 7^x=-2401:(-7)$

$\Rightarrow 7^x=343$

$\Rightarrow 7^x=7^3$

$\Rightarrow x=3$ (tmdk)

c) $\frac{625}{(-5)^x}=25$

$\Rightarrow (-5)^x=625:25$

$\Rightarrow (-5)^x=25$

$\Rightarrow (-5)^x=(-5)^2$

$\Rightarrow x=2$ (tmdk)

a: \(\left(\dfrac{2}{3}\right)^6\cdot\left(\dfrac{8}{27}\right)^2=\left(\dfrac{2}{3}\right)^6\cdot\left(\dfrac{2}{3}\right)^6=\left(\dfrac{2}{3}\right)^{12}\)

b: \(\left(\dfrac{3}{5}\right)^2\cdot\left(-\dfrac{9}{25}\right)^2=\left(\dfrac{3}{5}\right)^2\cdot\left(\dfrac{9}{25}\right)^2\)

\(=\left(\dfrac{3}{5}\right)^2\cdot\left(\dfrac{3}{5}\right)^4=\left(\dfrac{3}{5}\right)^6\)

c: \(\left(-\dfrac{5}{2}\right)^3:\left(-\dfrac{8}{125}\right)^3\)

\(=\left(-\dfrac{5}{2}:\dfrac{-8}{125}\right)^3=\left(\dfrac{5}{2}\cdot\dfrac{125}{8}\right)^3=\left(\dfrac{625}{16}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{12}\)