Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a) ĐK: x >= 0
<=> 7/2 + √x + 3/2 = 18
<=> 5 + √x = 18
<=> √x = 13
<=> x = 132 = 169 (nhận)
Vậy x = 169
b) ĐK: x >= 3
<=> √(x - 3) = 17
<=> x - 3 = 172 = 289
<=> x =292 (nhận)
Vậy x = 292

\(x:y:z=3:4:5\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\) và \(5z^2-3x^2-2y^2\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{5z^2-3x^2-2y^2}{5.5^2-3.3^2-2.4^2}=\frac{594}{66}=9\)
\(\Leftrightarrow\frac{x}{3}=9\Rightarrow x=9.3=27\)
\(\Leftrightarrow\frac{y}{4}=9\Rightarrow y=9.4=36\)
\(\Leftrightarrow\frac{z}{5}=9\Rightarrow z=9.5=45\)
Vậy x = 27 ; y = 36 ; z = 45
\(x+y=3\left(x-y\right)\)
\(\Rightarrow x+y=3x-3y\)
\(\Rightarrow y+3y=3x-x\)
\(\Rightarrow4y=2x\)
\(\Rightarrow2y=x\)
\(\Rightarrow x:y=2\)
\(\Rightarrow x+y=2y+y=2\)
\(\Rightarrow3y=2\)
\(\Rightarrow y=\frac{2}{3}\)
\(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3};y=\frac{2}{3}\)

a) Ta thấy:
\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{2}\cdot\frac{3}{5}=\frac{y}{3}\cdot\frac{3}{5}\)\(\Rightarrow\frac{3x}{10}=\frac{y}{5}\)
Mà \(\frac{y}{5}=\frac{z}{6}\) nên ta có biểu thức: \(\frac{3x}{10}=\frac{y}{5}=\frac{z}{6}\) ( 1 )
Biểu thức ( 1 ) tương đương với:
\(\frac{3x}{10}=\frac{3y}{15}=\frac{3z}{18}=\frac{3x+3y+3z}{10+15+18}=\frac{3\left(x+y+z\right)}{43}=\frac{3\cdot43}{43}=3\)
Khi đó:
\(\frac{3x}{10}=3\) \(\Rightarrow x=\frac{3\cdot10}{3}=10\)
\(\frac{3y}{15}=3\)\(\Rightarrow\frac{y}{5}=3\) \(\Rightarrow y=3\cdot5=15\)
\(\frac{3z}{18}=3\)\(\Rightarrow\frac{z}{6}=3\) \(\Rightarrow z=3\cdot6=18\)
a, Nhân cả hai vế cho 5, ta được: X/10 = Y/15
Tương tự ta có: Y/15 = Z/18
Do đó: X/10 = Z/18 (=Y/15)
Theo đề bài, ta có: (X+Y+Z)/(10+15+18) = 43/43 = 1
X/10=1 => X=10
Y/15=1 => Y=15
Z/18=1 => Z=18

\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^{10}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{30}\)
\(\Leftrightarrow2x=30\Leftrightarrow x=15\)
Bài làm
Ta có: \(\left(\frac{4}{9}\right)^x=\left(\frac{2}{3}\right)^{2x}\)
\(\left(\frac{8}{27}\right)^{10}=\left(\frac{2}{3}\right)^{3.10}=\left(\frac{2}{3}\right)^{30}\)
\(\Rightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{30}\)
\(\Rightarrow2x=30\)
\(\Rightarrow x=15\)
Vậy \(x=15\)
# Học tốt #

- \(x+y=\frac{7}{12}\Rightarrow x=\frac{7}{12}-y\)(1)
- \(y+z=\frac{-19}{24}\Rightarrow z=\frac{-19}{24}-y\)(2)
- \(z+x=\frac{1}{8}\)(3)
Từ (1),(2) và (3) \(\Rightarrow\left(\frac{-19}{24}-y\right)+\left(\frac{7}{12}-y\right)=\frac{1}{8}\)
\(\Rightarrow\frac{-19}{24}-y+\frac{7}{12}-y=\frac{1}{8}\)
\(\Rightarrow\frac{-5}{24}-2y=\frac{1}{8}\)
\(\Rightarrow-2y=\frac{1}{8}+\frac{5}{24}\)
\(\Rightarrow-2y=\frac{1}{3}\)
\(\Rightarrow y=\frac{1}{3}:\left(-2\right)\)
\(\Rightarrow y=-\frac{1}{6}\)
Vì \(x=\frac{7}{12}-y\)mà \(y=-\frac{1}{6}\);\(\Rightarrow x=\frac{7}{12}-\frac{-1}{6}=\frac{7}{12}+\frac{1}{6}=\frac{7+2}{12}=\frac{9}{12}=\frac{3}{4}\)
Vì \(z=\frac{-19}{24}-y\)mà \(y=-\frac{1}{6}\);\(\Rightarrow z=\frac{-19}{24}-\frac{-1}{6}=\frac{-19}{24}+\frac{1}{6}=\frac{-19+4}{24}=\frac{-15}{24}=\frac{-5}{8}\)
Vậy
- \(x=\frac{3}{4}\)
- \(y=-\frac{1}{6}\)
- \(z=-\frac{5}{8}\)

\(=\frac{8}{9}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{8}{9}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-\frac{1}{6}+\frac{1}{7}-\frac{1}{5}+\frac{1}{6}-...-1+\frac{1}{2}\)= 0
Vì \(\frac{1}{n.\left(n+1\right)}=\frac{\left(n+1\right)-n}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)