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Đề tham khảo nè chị j

#Châu's ngốc

xét tứ giác AFCD có EA=EC;ED=EF nên tứ giác AFCD là hình bình hành

\(P=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2+\left(z+\dfrac{1}{z}\right)^2-\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\left(z+\dfrac{1}{z}\right)\) 

Ta có: \(xyz=1\Rightarrow x=\dfrac{1}{yz}\) 

\(P=\left(\dfrac{1}{yz}+yz\right)^2+\left(y+\dfrac{1}{y}\right)^2+\left(z+\dfrac{1}{z}\right)^2-\left(yz+\dfrac{1}{yz}\right)\left(y+\dfrac{1}{y}\right)\left(z+\dfrac{1}{z}\right)\)

\(P=\dfrac{1}{y^2z^2}+2+1y^2z^2+y^2+2+\dfrac{1}{y^2}+z^2+2+\dfrac{1}{z^2}-\left(y^2z+z+\dfrac{1}{z}+\dfrac{1}{y^2z}\right)\left(z+\dfrac{1}{z}\right)\)

\(P=\dfrac{1}{y^2z^2}+y^2z^2+y^2+\dfrac{1}{y^2}+z^2+\dfrac{1}{z^2}+6-y^2z^2-y^2-z^2-1-1-\dfrac{1}{z^2}-\dfrac{1}{y^2}-\dfrac{1}{y^2z^2}\)\(P=\left(\dfrac{1}{y^2z^2}-\dfrac{1}{y^2z^2}\right)+\left(y^2z^2-y^2z^2\right)+\left(y^2-y^2\right)+\left(z^2-z^2\right)+\left(\dfrac{1}{y^2}-\dfrac{1}{y^2}\right)+\left(\dfrac{1}{z^2}-\dfrac{1}{z^2}\right)+4\)

\(P=4\)

Vậy: ... 

\(\frac{8}{56}+\frac{8}{140}+\frac{8}{260}+...+\frac{8}{1400}=\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...+\frac{4}{700}\)

\(=\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}\)

\(=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{4}{3}.\frac{3}{14}=\frac{2}{7}\)

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

(3⁸ . 3¹⁶) : (3 ⁷ . 3¹⁴)

3²⁴ : 3²¹

3³

HK TỐT

(3⁸ . 3¹⁶) : (3⁷ . 3¹⁴) = 3⁸⁺¹⁶ : 3⁷⁺¹⁴

                                 = 3²⁴ : 3²¹

                                 = 3^24-21

                                 = 3³ = 27

Trả lời :

(3⁸ . 3¹⁶) : (3⁷ . 3¹⁴)

= 3^{8 + 16} : 3^{7 + 14}

= 3²⁴ : 3²¹

= 3³

= 27

= 3³⁰

\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)

\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)

\(=0\)

Ta có :                  

                ^{82 - 576 : 32}

^{= 64 - 576 : 9}

^{= 64 - 64}

^{=  0}

 (1¹ + 2² + 3³ + 4⁴ +...+ 2022²⁰²²) . 0

^{= 0}           

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x=t\)

\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)

\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)

Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

CHAO BAN

1+1bang may