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a)\(\frac{x-1}{x-5}=\frac{6}{7}\) điều kiện : x khác 5
<=>7x-7=6x-30<=> x=-23
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
<=> -60+35x=-52+39x
<=> 4x=-8
<=> x=-2
Ta có:
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{z+x}{zx}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)
Thay tất cả giá trị x,y,z vào M ta được:
\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)
\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)
\(\Rightarrow M=2020+2021=4041\)
Ta có: x=2018
nên x+1=2019
Ta có: \(A=x^5-2019x^4+2019x^3-2019x^2+2019x-2020\)
\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2020\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2020\)
\(=x-2020=2019-2020=-1\)
\(\left(x-5\right)^{2020}+\left(y-x+1\right)^{2022}=0\left(1\right)\)
Ta có \(\left\{{}\begin{matrix}\left(x-5\right)^{2020}\ge0,\forall x\\\left(y-x+1\right)^{2022}\ge0,\forall x;y\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^{2020}=0\\\left(y-x+1\right)^{2022}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-5=0\\y-x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y-5+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)
( 2x - 5 )2020 + ( 5y + 1 )2022 ≤ 0
Ta có : ( 2x - 5 )2020 ≥ 0 ∀ x
( 5y + 1 )2022 ≥ 0 ∀ y
=> ( 2x - 5 )2 + ( 5y + 1 )2022 ≥ 0 ∀ x, y
Kết hợp với đề bài => Chỉ xảy ra trường hợp ( 2x - 5 )2020 + ( 5y + 1 )2022 = 0
Khi đó \hept{2�−5=05�+1=0⇔\hept{�=52�=−15\hept{2x−5=05y+1=0⇔\hept{x=25y=−51
a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x-1}{3}=\frac{y-1}{4}=\frac{z+2}{5}=\frac{z-1+y-1+z+2}{3+4+5}=\frac{-36}{12}=-3\)
=> \(\hept{\begin{cases}\frac{x-1}{3}=-3\\\frac{y-1}{4}=-3\\\frac{z+2}{5}=-3\end{cases}}\) => \(\hept{\begin{cases}x-1=-9\\y-1=-12\\z+2=-15\end{cases}}\) => \(\hept{\begin{cases}x=-8\\x=-11\\x=-13\end{cases}}\)
Vậy ...
tui làm ròi mà :>
\(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)
Ta có: \(\left(3x-\frac{1}{5}\right)^{2020}\ge0\forall x\)
\(\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}\ge0\forall y\)
Mà \(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-\frac{1}{5}\right)^{2020}=0\\\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-\frac{1}{5}=0\\\frac{2}{5}\cdot y+\frac{4}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=\frac{1}{5}\\\frac{2}{5}\cdot y=\frac{-4}{7}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{15}\\y=\frac{-10}{7}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\frac{1}{15};\frac{-10}{7}\right)\)