a) Ta có: \(C=-\left|x+2\right|\le0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x+2\right|=0\Rightarrow x=-2\)

Vậy Max(C) = 0 khi x = -2

b) Ta có: \(D=1-\left|2x-3\right|\le1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|2x-3\right|=0\Rightarrow x=\frac{3}{2}\)

Vậy Max(D) = 1 khi x = 3/2

d) \(D=-\left|x+\frac{5}{2}\right|\le0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left|x+\frac{5}{2}\right|=0\Rightarrow x=-\frac{5}{2}\)

Vậy Max(D) = 0 khi x = -5/2

e) \(P=4-\left|5x-3\right|-\left|3y+12\right|\le4\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|5x-3\right|=0\\\left|3y+12\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)

Vậy Max(P) = 4 khi \(\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)

Đặt A = 1.4 + 2.5 + 3.6 + ... + 100.103

= 1.(2 + 2) + 2.(3 + 2) + 3.(4 + 2) +.... + 100.(101 + 2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + (1.2 + 2.2 + 3.2 + ... + 100.2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 2(1 + 2 + 3 + .... + 100)

= 1.2 + 2.3 + 3.4 + .... + 100.101 + 2.100.(100 + 1) : 2

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 10100

Đặt B = 1.2 + 2.3 + 3.4 + .... + 100.101

=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + 100.101.3

=> 3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)

=> 3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101

=> 3B = 100.101.102

=> B = 343400

Khi đó A = 343400 - 10100 = 333300

bạn tính kiểu khác đc ko ? kiểu ab mình ko hiểu lắm

T=(12−13)(12−15)(12−17).....(12−199)T=(12−13)(12−15)(12−17).....(12−199)

⟹T=12(1−23).12(1−25).12(1−27).....12(1−299)⟹T=12(1−23).12(1−25).12(1−27).....12(1−299)

Thấy T có: (99-3):2+1=49(SH)

⟹T=(12.49).[(1−23).(1−25)...(1−299)⟹T=(12.49).[(1−23).(1−25)...(1−299)

⟹T=492.199=49198

DE SAI ROI EM EI

Để A đạt GTNN

=> \(\frac{1}{3,5-\left|x+5\right|}\)đạt GTNN

=> 3,5 - |x + 5| đạt GTLN (ĐK 3,5 -  |x + 5| \(\ne\)0)

mà \(\left|x+5\right|\ge0\forall x\Rightarrow3,5-\left|x+5\right|\le3,5\)

Dấu "=" xảy ra <=> x + 5 = 0 => x = -5

=> 3,5 - |x + 5| đạt GTLN là 3,5 <=> x = -5

Thay x vào A 

=> GTNN của A LÀ 1/3,5 <=> x = -5

a) \(\left|x-1,7\right|=2,3\)

\(\Leftrightarrow\orbr{\begin{cases}x-1,7=2,3\\x-1,7=-2,3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-0,6\end{cases}}\)

b) \(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=-\frac{1}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{5}{12}\\x=-\frac{13}{12}\end{cases}}\)

c) \(\left|x+\frac{1}{4}\right|-\frac{3}{4}=0\)

\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{3}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\frac{3}{4}\\x+\frac{1}{4}=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)

d) \(2-\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{4}\)

\(\Leftrightarrow\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{3}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x-\frac{1}{4}=\frac{3}{4}\\\frac{3}{2}x-\frac{1}{4}=-\frac{3}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{3}{2}x=1\\\frac{3}{2}x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{1}{3}\end{cases}}\)

e) \(\left|4+2x\right|+4x=0\)

\(\Leftrightarrow\left|4+2x\right|=-4x\)

\(\Leftrightarrow\orbr{\begin{cases}4+2x=-4x\\4+2x=4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}-6x=4\\2x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)

a) Đề chắc là: \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)

Ta có: \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\left(\forall x,y,z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{378}{395}\\z=2004\end{cases}}\)

b) Ta có: \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\left(\forall x,y,z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}\)

a) Vì : \(\left|x+\frac{19}{5}\right|\ge0\forall x\in R\)

\(\left|y+\frac{1890}{1975}\right|\ge0\forall y\in R\)

\(\left|z-2004\right|\ge0\forall z\in R\)

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\in R\)

Dấu''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-1890}{1975}\\z=2004\end{cases}}\)

b,\(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

Ta có:\(\left|x+\frac{9}{2}\right|\ge0\forall x\)

          \( \left|y+\frac{4}{3}\right|\ge0\forall y\)

          \(\left|z+\frac{7}{2}\right|\ge0\forall z\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\forall x,y,z\)

Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}}\)

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hỏi chụy Google

Ta có : AB là đường trung trực của MN

             CD là đường trung trực của MP

              EF là đường trung trực của NP