\(e.\left(x+\frac{2019}{2020}\right)^{100}+\left(y-\frac{9}{11}\right)^{200}=0\)
Ai làm nhanh mik tim cho
\(a.\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
Ai làm nhanh mik tim cho
\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
Vì \(\left|x-\frac{2}{3}\right|\ge0\)và \(\left|y+\frac{5}{9}\right|\ge0\)nên \(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|\ge0\)
(Dấu "="\(\Leftrightarrow\)\(\left|x-\frac{2}{3}\right|=0\)và \(\left|y+\frac{5}{9}\right|=0\))
\(\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}\)
vì \(\left|x-\frac{2}{3}\right|>0\)hoặc =0 ;\(\left|y+\frac{5}{9}\right|>0\)hoặc =o
mà\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
nên |x-2/3| =0 và |y+5/9|=0
\(\Rightarrow\hept{\begin{cases}x-\frac{2}{3}=0\\y+\frac{5}{9}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}}\)
Ta có : \(\left|x-\frac{2}{3}\right|\ge0\forall x\)
\(\left|y+\frac{5}{9}\right|\ge0\forall y\)
\(\Leftrightarrow\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|\ge0\forall x,y\)
Dấu " = " xảy ra khi : \(\hept{\begin{cases}x-\frac{2}{3}=0\\y+\frac{5}{9}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{5}{9}\end{cases}}\)
Vậy : ...
Tìm GTNN
a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)
b. \(B=\left(x-\frac{5}{4}\right)^{20}+\left(y+\frac{4}{3}\right)^{30}-11\)
AI LÀM NHANH TỚ TIM
a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)
Ta có: \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\)
Dấu = xảy ra khi :
\(\frac{x}{5}+\frac{23}{2}=0\Leftrightarrow\frac{x}{5}=-\frac{23}{2}\Leftrightarrow x=-\frac{115}{2}\)
\(y-\frac{14}{3}=0\Leftrightarrow y=\frac{14}{3}\)
Vậy ..............
Ta có:
a) \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall y\)
=> \(\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{5}+\frac{23}{2}=0\\y-\frac{14}{3}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
Vậy Min của A = 2019 tại \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
câu b tượng tự
\(b,B=\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{30}-11\)
Ta có : \(\left[x-\frac{5}{4}\right]^{20}\ge0\forall x\)
\(\left[y-\frac{4}{3}\right]^{30}\ge0\forall y\)
\(\Leftrightarrow\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{20}-11\ge-11\forall x,y\)
Dấu " = " xảy ra : \(\hept{\begin{cases}\left[x-\frac{5}{4}\right]^{20}=0\\\left[y-\frac{4}{3}\right]^{20}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\frac{5}{4}=0\\y-\frac{4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=\frac{4}{3}\end{cases}}\)
Vậy : ...
Tìm x biết \(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)}\)\(\frac{+\left(x-2020\right)^2}{+\left(x-2020\right)^2}\)\(=\frac{19}{49}\)
a,Cho \(\left(x-2019+\sqrt{\left(x-2019\right)^2+2020}\right)\left(y-2019+\sqrt{\left(y-2019\right)^2+2020}\right)=2020\)Tính : D = x + y
b, Cho \(\frac{-3}{2}\le x\le\frac{3}{2},x\ne0,a=\sqrt{3+2x}-\sqrt{3-2x}\)
Tính : \(G=\frac{\sqrt{6+2\sqrt{9-4x^2}}}{x}\) theo a.
Em cảm ơn mọi người nhiều ạ.
tìm x biết
\(\frac{\left(2019-x^2\right)+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020^2\right)}\) = \(\frac{19}{49}\)
Tìm x, biết:
\(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}=\frac{19}{49}\)
Các bạn mong giúp mình sớm nhé
ủa bạn j ơi chữ x chành bành ra trên đề kìa mà bạn bảo tìm làm j nữa
Cho hàm số \(f\left(x\right)=\frac{100^x}{100^x+10}\)
a, Chứng minh rằng nếu a,b là 2 số thỏa mãn a + b = 1 thì f(a) + f(b) = 1
b,Tính tổng \(A=f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+...+f\left(\frac{2019}{2020}\right)\)
Cho hàm số \(f\left(x\right)=\frac{100^x}{100^x+10}\)
a, Chứng minh rằng nếu a,b là 2 số thỏa mãn a + b = 1 thì f(a) + f(b) = 1
b,Tính tổng \(A=f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+...+f\left(\frac{2019}{2020}\right)\)
Tìm x khi biết x:☺
\(\left(\frac{x-4}{2017}\right)+\left(\frac{x-3}{2018}\right)+\left(\frac{x-2}{2019}\right)+\left(\frac{x-1}{2020}\right)=4\)
gip mik với mik cần gấppppppppppppppppp
\(\frac{x-4}{2017}+\frac{x-3}{2018}+\frac{x-2}{2019}+\frac{x-1}{2020}=4\\ \Leftrightarrow\left(\frac{x-4}{2017}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-2}{2019}-1\right)+\left(\frac{x-1}{2020}-1\right)=4-1-1-1\)
\(\Leftrightarrow\frac{x-2021}{2017}+\frac{x-2021}{2018}+\frac{x-2021}{2019}+\frac{x-2021}{2020}=0\)
\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2021=0\\\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\end{matrix}\right.\)
\(\Leftrightarrow x=2021\)
Vậy...