h2o+ h2o = bao nhiêu

Nếu bài tìm giá trị lớn nhất , ta làm như sau :

Vì | -1,8 + x | ≥ 0 ∀ x 

=> -3,1 - | -1,8 + x | ≤ -3,1 ∀ x

=> A ≤ -3,1

Dấu "=" xảy ra <=> | -1,8 + x | = 0

<=> -1,8 + x = 0

<=> x = 1,8

Vậy giá trị lớn nhất của A = -3,1 <=> x = 1,8

\(=\left(-\frac{37}{38}\right)\left(-\frac{36}{37}\right)\left(-\frac{35}{36}\right)...\left(-\frac{31}{32}\right)=\left(-\frac{31}{38}\right)\)

a. 0.7777= 0.(7)

b. -5.123123123...= - 5.(123)

c. 4.7513513513...=4.7(513)

d. - 17.32405405405... = - 17.32(405)

a) Vì \(\hept{\begin{cases}\left|x-y-2\right|\ge0∀x,y\\\left|y+3\right|\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-y-2\right|+\left|y+3\right|≥0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-y-2\right|=0\\\left|y+3\right|=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-y-2=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=2\\y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-3\end{cases}}\)

b) Vì \(\hept{\begin{cases}\left|x-3y\right|^{2007}\ge0∀x,y\\\left|y+4\right|^{2008}\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-3y\right|^{2007}+\left|y+4\right|^{2008}≥0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-3y\right|^{2007}=0\\\left|y+4\right|^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x-3y\right|=0\\\left|y+4\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3y=0\\y+4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3y\\y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-12\\y=-4\end{cases}}\)

c) Vì \(\hept{\begin{cases}\left(x+y\right)^{2006}\ge0∀x,y\\2007.\left|y-1\right|\ge0∀y\end{cases}}\)\(\Rightarrow\left(x+y\right)^{2006}+2007.\left|y-1\right|\ge0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+y\right)^{2006}=0\\2007.\left|y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\\left|y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-y\\y-1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

d) Vì \(\hept{\begin{cases}\left|x-y-5\right|\ge0∀x,y\\2007.\left(y-3\right)^{2008}\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-y-5\right|+2007.\left(y-3\right)^{2008}\ge0\ge0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-y-5\right|=0\\2007.\left(y-3\right)^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y-5=0\\\left(y-3\right)^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=5\\y-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y=5\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)

hello

-33/19 < 45/-31

nhớ tít nha

Ta có :

( x - 1 ) . ( x - 1 ) . ( x - 1 ) . ( x - 1 ) = ( x - 1 ) . ( x - 1 )

=> ( x - 1 )⁴ = ( x - 1 )²

=> ( x - 1 )⁴ - ( x - 2 )² = 0

=> ( x - 1 )² . [ ( x - 1 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1\in\left\{-1;1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x\in\left\{0;2\right\}\end{cases}}\)