Cách 1 : \(\left(x-1\right)^{2019}+\left(x-1\right)^{2020}=0\)

Vì \(\left(x-1\right)^{2019}\ge0\forall x;\left(x-1\right)^{2020}\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^{2019}+\left(x-1\right)^{2020}\ge0\forall x\)

Dấu ''='' xảy ra <=> x = 1

Cách 2 : \(\left(x-1\right)^{2019}+\left(x-1\right)^{2020}=0\)

\(\Leftrightarrow\left(x-1\right)^{2019}\left[1+\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{2019}=0\Leftrightarrow x=0;1\)

ok, bn...đáp án là... mk éo bt làm :))

@Khuê =)) vậy là bn ''gà'' hơn mk rồi.

\(f\left(x\right)=7\)hay \(9x^2-2=7\)

\(\Leftrightarrow9x^2=9\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)

\(f\left(x\right)=1\)hay \(9x^2-2=1\Leftrightarrow x^2=\frac{1}{3}\Leftrightarrow x=\pm\sqrt{\frac{1}{3}}\)

a) Ta có \(\frac{a+b-3c}{c}=\frac{b+c-3a}{a}=\frac{c+a-3b}{b}\)

=> \(\frac{a+b-3c}{c}+4=\frac{b+c-3a}{a}+4=\frac{c+a-3b}{b}+4\)

=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

Vì a;b;c > 0

=> a + b + c > 0

=> \(\frac{1}{c}=\frac{1}{a}=\frac{1}{b}\)=> a = b = c (đpcm)

2) Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)

=> \(\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\)

Khi đó M = 4(a - b)(b - c) - (c - a)²

= 4(2016k - 2017k)(2017k - 2018k) - (2018k - 2016k)²

= 4(-k).(-k) - (2k)²

= 4k² - 4k² = 0

\(\frac{4}{11.16}+\frac{4}{16.21}+\frac{4}{21.26}+...+\frac{4}{61.66}\)

= \(\frac{4}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\right)\)

\(=\frac{4}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(=\frac{4}{5}\left(\frac{1}{11}-\frac{1}{66}\right)=\frac{4}{5}.\frac{5}{66}=\frac{2}{33}\)

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)

b) Ta có A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2\left(2n+1\right)}< \frac{1}{2}\)(đpcm)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}=\frac{y}{4}=\frac{x-2y}{3-2.4}=\frac{10}{3-8}=\frac{10}{-5}=-2\\ \frac{x}{3}=-2\Rightarrow x=-2.3=-6\\ \frac{y}{4}=-2\Rightarrow y=-2.4=-8\)

Vậy x=-6

y=-8
 

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{3}=\frac{y}{4}=\frac{x-2y}{3-8}=-\frac{10}{-5}=2\)

\(x=6;y=8\)