Từ \(x\ge2\) cộng cả hai vế với \(\dfrac{1}{2}\) ta được

\(x+\dfrac{1}{2}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(VT=x+\dfrac{1}{2}=x-2+2+\dfrac{1}{2}=\left(x-2\right)+\dfrac{5}{2}\)

\(\left\{{}\begin{matrix}x\ge2\Rightarrow x-2\ge0\\VT=\left(x-2\right)+\dfrac{5}{2}\ge\dfrac{5}{2}=VP\rightarrow dpcm\end{matrix}\right.\)

   18/37 + 8/24 + 19/37 - 1/23/24 + 2/3

= 18/37 + 19/37 +  1/3 + 2/3 - 1/23/24

= 1 + 1 - 1/23/24

= 2 - 1/23/24

= 1/24

( chú ý : 1/3 là rút gọn của 8/24 )

\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)

\(=\frac{18}{37}+\frac{1}{3}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)

\(=\left(\frac{18}{37}+\frac{19}{37}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)-1\frac{23}{24}\)

\(=1+1-1\frac{23}{24}\)

\(=2-1\frac{23}{24}=\frac{1}{24}\)

\(\frac{18}{27}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)

\(=\frac{18}{27}+\frac{8}{24}+\frac{19}{37}-\frac{47}{24}+\frac{16}{24}\)

\(=\left(\frac{18}{37}+\frac{19}{37}\right)+\left(\frac{8}{24}+\frac{16}{24}\right)-\frac{47}{24}\)

\(=\frac{37}{37}+\frac{24}{24}-\frac{47}{24}\)

\(=1+1-\frac{47}{24}\)

\(=2-\frac{47}{24}\)

\(=\frac{48}{24}-\frac{47}{24}\)

\(=\frac{1}{24}\)

a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)

Vậy ...

trả lời

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

hk tốt

b) ta có: \(5^{50}=\left(5^2\right)^{25}=25^{25}\); \(3^{75}=\left(3^3\right)^{25}=27^{25}\).

Mà \(25^{25}< 27^{25}\)nên \(5^{50}< 3^{75}\)

1) x(x-2) + 3(x+5) + 4x -15 =0

=> x\(^2\) - 2x + 3x + 15 + 4x - 15 = 0

=> ( x\(^2\) -2x + 3x + 4x ) + 15 - 15 = 0

=> x \(^2\) -2x+3x+4x = 0

=> x(x-2+3+4)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2+3+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)

2) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=2017\)

\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017.2017\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=2017^2\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=2017^2\)

\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{c}{a+b}\right)=2017^2\)

\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{c}{a+b}\right)=2017^2\)

\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)

xin lỗi mik xin đc sửa lại 3 dòng cuối vì mik ghi nhầm :

\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2017^2\)

\(\Rightarrow\left(1+\frac{c}{a+b}\right)+\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)=2017^2\)

\(\Rightarrow3+\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=2017^2\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2017^2-3\)

khong biet

x=4, y=6,z=8

Trần Thảo Trang

Áp dụng tính chất dãy tỉ số bằng nhau,ta có: 
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{18}{9}=2\)

\(\frac{x}{2}=2\Rightarrow x=2.2=4\)

\(\frac{y}{3}=2\Rightarrow y=2.3=6\)

\(\frac{z}{4}=2\Rightarrow z=2.4=8\)

Vậy x = 4, y = 6 và z = 8