A = -\(x^2\) -  0,75 

\(x^2\) ≥ 0 ∀ \(x\) ⇒ -\(x^2\) ≤ 0 ⇒ - \(x^2\) - 0,75 ≤ -0,75  

A_max = -0,75 ⇔ \(x\) = 0

Do x² ≥ 0 với mọi x ∈ R

⇒ -x² ≤ 0 với mọi x ∈ R

⇒ -x² - 0,75 ≤ -0,75 với mọi x ∈ R

Vậy GTLN của A là -0,75 khi x = 0

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

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

Vậy giá trị cần tìm là 972

\(=\dfrac{\left(2^2\cdot3\right)^3\cdot\left(2\cdot3^2\right)^2}{\left(2^3\cdot3\right)^2}\)
\(=\dfrac{2^{2\cdot3}\cdot3^3\cdot2^2\cdot3^{2\cdot2}}{2^{3\cdot2}\cdot3^2}\)
\(=\dfrac{2^6\cdot2^2\cdot3^3\cdot3^4}{2^6\cdot3^2}\)
\(=2^2\cdot3^5\) \(=4\cdot243\) \(=486\)

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

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

= 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.

Nghi viết sai đề :))

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

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

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 \(A< \dfrac{-11}{21}\)

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

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

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

= \(\dfrac{-4}{13}.\dfrac{17}{17}\)

= \(\dfrac{-4}{13}.1\)

= \(\dfrac{-4}{13}\)

= \(\dfrac{-4.5-12.4}{13.17}\)

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

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

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

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

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

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: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989A=(4⋅91​+9⋅141​+14⋅191​+...+44⋅491​)⋅891−3−5−7−...−49​

⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅95​+9⋅145​+14⋅195​+...+44⋅495​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−91​+91​−141​+141​−191​+...+441​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅4949−4​)⋅891−3−5−7−...−49​

⇔�=15⋅45196⋅1−3−5−7−...−4989⇔A=51​⋅19645​⋅891−3−5−7−...−49​

⇔�=9196⋅1−3−5−7−...−4989⇔A=1969​⋅891−3−5−7−...−49​

⇔�=9196⋅−62389=−928⇔A=1969​⋅89−623​=−289​
 

A = -9/28

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

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

viết đề sai kìa bạn tui ới.

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

Min-max tưởng bạn học từ lớp 7 rồi.

Min : Gía trị nhỏ nhất

Max: Gía trị lớn nhất