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\(\sqrt{x-1}=3.\) \(\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
Ta có công thức : \(\sqrt{x-1}^2=n^2\) thì mới phá được dấu căn bậc 2
Nên ta làm như sau :
\(\sqrt{x-1}=3.\) \(\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
\(\sqrt{x-1}=3.\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=3\)
\(\sqrt{x-1}^2=3^2\)
\(x-1=9\)
\(x=9+1\)
\(\Rightarrow x=10\)
\(\text{Ta có:}\left(x+2019\right)^{2018}\ge0với\forall x\)
\(|y-2020|\ge0với\forall y\)
\(\Rightarrow\)\(\left(x+2019\right)^{2018}+\)\(|y-2020|\ge0với\forall x,y\)
\(\text{Mà }\)\(\left(x+2019\right)^{2018}+\)\(|y-2020|=0\)\(\text{(Theo đề bài)}\)
\(\Rightarrow\hept{\begin{cases}\left(x+2019\right)^{2018}=0\\|y-2020|=0\end{cases}\Rightarrow\hept{\begin{cases}x+2019=0\\y-2020=0\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=-2019\\y=2020\end{cases}}\)
\(\Rightarrow M=x+y=-2019+2020=1\)
Ta có : (7x - 5y)2018 + (3x - 2z)2020 + (xy + yz + xz - 4500)2018 = 0
Ta có : \(\hept{\begin{cases}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+xz-4500\right)^{2018}\ge0\end{cases}}\)
\(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+xz-4500\right)^{2018}\ge0\)
Dấu bằng xảy ra <=>
\(\begin{cases}7x=5y\\3x=2z\\xy+yz+xz=4500\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+xz=4500\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+xz=4500\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\x+y+z=4500\end{cases}}\)
Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\hept{\begin{cases}x=10k\\y=14k\\z=15k\end{cases}}\)
=> xy + yz + xz = 4500
<=> 10k.14k + 14k.15k + 10k.15k = 4500
=> 140.k2 + 210.k2 + 150.k2 = 4500
=> k2.(140 + 210 + 150) = 4500
=> k2 . 500 = 4500
=> k2 = 9
=> k = \(\pm3\)
Nếu k = 3
=> \(\hept{\begin{cases}x=30\\y=42\\z=45\end{cases}}\)
Nếu k = - 3
=> \(\hept{\begin{cases}x=-30\\y=-42\\z=-45\end{cases}}\)
(x-2020)x - 1 - (x - 2020)x + 2019 = 0
=> (x - 2020)x - 1 .[(x - 2020)2020 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-2020\right)^{x-1}=0\\\left(x-2020\right)^{2020}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-2020=0\\\left(x-2020\right)^{2020}=1^{2020}\end{cases}\Rightarrow}\orbr{\begin{cases}x-2020=0\\x-2020=\pm1\end{cases}}}\)
=> \(x-2020\in\left\{0;1;-1\right\}\Rightarrow x\in\left\{2020;2021;2019\right\}\)
\(\left(x-1\right)^4=\left(1-x\right)^6\Leftrightarrow\left(x-1\right)^4=\left(x-1\right)^6\)
\(\Leftrightarrow\left(x-1\right)^4\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^4=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
a, (x-1)4=(1-x)6
⇒ (x-1)4=(x-1)6
⇒ (x-1)4 - (x-1)6 =0
⇒ (x-1)4 (1-(x-1)6)=0
⇒ \(\left[{}\begin{matrix}\left(x-1\right)^4=0\\1-\left(x-1\right)^6=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^6=1\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=1\\x-6=1\\x-6=-1\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=1\\x=7\\x=5\end{matrix}\right.\)
Vậy x ∈ \(\left\{1;7;5\right\}\)
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
\(\sqrt{x-1}=5.\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}^2=5^2\)
\(x-1=25\)
\(x=25+1\)
\(\Rightarrow x=26\)
Mình làm hơi tắt, để mình làm lại nhé!
\(\sqrt{x-1}=5.\left(2018+2019+2020\right)^0\)
\(\sqrt{x-1}=5\)
\(\sqrt{x-1}^2=5^2\)
\(x-1=25\)
\(x=25+1\)
\(\Rightarrow x=26\)