a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)

b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)

TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)

TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)

Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)

a,\(\frac{3}{4}-x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)

b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)

\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)

TH₁:\(x+\frac{2}{3}=\frac{5}{6}\)

\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)

\(\Leftrightarrow x=\frac{1}{6}\)

TH₂:\(x+\frac{2}{3}=-\frac{5}{6}\)

\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{3}{2}\)

Cho hỏi D ở đâu vậy

đánh nhầm bạn ạ

có trừ hết bài ko ạ

a, \(\frac{x+5}{x+3}< 1\Leftrightarrow x+5< x+3\)

\(\Leftrightarrow5< 3\)( vô lí vaichuong =)) 

Vậy ko có giá trị x thỏa mãn đề bài 

b, \(\frac{x+3}{x+4}>1\Leftrightarrow x+3>x+4\)

\(\Leftrightarrow3>4\)( vô lí ) 

Vậy ko có giá trị x thỏa mãn đề bài 

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

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

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

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

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