\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)

Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5

\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)

Ta có: $A=\left(\genfrac{}{}{0ex}{}{1}{4\cdot 9}+\genfrac{}{}{0ex}{}{1}{9\cdot 14}+\genfrac{}{}{0ex}{}{1}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{1}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{5}{4\cdot 9}+\genfrac{}{}{0ex}{}{5}{9\cdot 14}+\genfrac{}{}{0ex}{}{5}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{5}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{9}+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{14}+\genfrac{}{}{0ex}{}{1}{14}-\genfrac{}{}{0ex}{}{1}{19}+...+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{49-4}{4\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{45}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{-623}{89}=-\genfrac{}{}{0ex}{}{9}{28}$
 

\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)

\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)

mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)

\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)

Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)

\(\Rightarrow A>-\dfrac{11}{21}\)

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)

\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)

\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)

Dễ thấy A có 9 thừa số, suy ra

\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)

Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)

Vậy \(A< \dfrac{-11}{21}\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)

\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)

\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)

\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)

\(\Leftrightarrow376x+752=375x+1125\)

\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)

\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{2^3.3^3+2.2^2.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\=\dfrac{2^3.3^3+2^3.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2^3.\left(3^3+3^2+1\right)}{37}=\dfrac{2^3.37}{37}=2^3=8\)

\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{216+72+8}{37}=\dfrac{296}{37}=8\)

\(\dfrac{12^3.18^2}{24^2}=\dfrac{12^3.6^2.3^2}{6^2.4^2}=\dfrac{4^3.3^3.3^2}{4^2}=4.3^3.3^2=4.3^5=972\)

\(\dfrac{12^3.18^2}{24^2}=\dfrac{1728.324}{576}=\dfrac{559872}{576}=972\)

Vậy giá trị cần tìm là 972

a) Ta có: x²\(\ge0,\forall x\) 

=> x² +3/4 \(\ge\dfrac{3}{4}\) , mọi x

Vậy min A = 3/4

Dấu "=" xảy ra <=> x =0

b) ( x- 3/2)² -0,4

Ta có ( x-3/2)² lớn hơn hoặc bằng 0, mọi x

=> ( x-3/2)² - 0,4 lớn hơn hoặc bằng 0 - 0;4 = -0,4

Vậy min B =-0,4

Dấu "=" xảy ra <=> x = 3/2

Chúc bạn học tốt !

bạn cho mik hỏi là min A nghĩa là sao vậy

   \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)

= \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)

= \(\dfrac{-4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)

= \(\dfrac{-4}{13}.\dfrac{17}{17}\)

= \(\dfrac{-4}{13}.1\)

= \(\dfrac{-4}{13}\)

= \(\dfrac{-4.5-12.4}{13.17}\)

=\(\dfrac{-4\left(5+12\right)}{13.17}\)

=\(\dfrac{-4.17}{13.17}\)

=\(\dfrac{-4}{13}\)

= 7 - 3 căn 5 + 1

= 

Nó ra xấp xỉ mà nhỉ đề vô lí vậy ta

Giải thích seo cho hs lớp 7 hiểu cùng ạ.