Tìm x
\(\frac{x}{4}=\frac{18}{x+1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x}{4}=\frac{18}{x+1}\Rightarrow x\left(x+1\right)=18\cdot4\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+9x-8x-72=0\)
\(\Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\)
\(\Rightarrow\left(x-8\right)\left(x+9\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=8\\x=-9\end{array}\right.\)
nhìn sơ là biết sai. x và x + 1 là hai số tn liên tiếp. mà 8 vs -9
a/ \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-1\right)=72\)
\(\Leftrightarrow x-1=24\)
\(\Leftrightarrow x=25\)
Vậy ..
b/ \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x^2=6^2=\left(-6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy ..
c/ \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x\left(x+1\right)=72\)
\(\Leftrightarrow x\left(x+1\right)=8.9\)
\(\Leftrightarrow x=8\)
Vậy ..
a, \(\frac{x}{y+z+1}=\frac{y}{x+z+3}=\frac{z}{x+y-4}=\frac{x+y+z}{y+z+1+x+z+3+x+y-4}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
=>\(x+y+z=\frac{1}{2};\frac{x}{y+z+1}=\frac{1}{2};\frac{y}{x+z+3}=\frac{1}{2};\frac{z}{x+y-4}=\frac{1}{2}\)
=>\(\hept{\begin{cases}y+z+1=2x\\x+z+3=2y\\x+y-4=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+3=3y\\x+y+z-4=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+3\\3z=\frac{1}{2}-4\end{cases}}}\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{7}{2}\\3z=\frac{-7}{2}\end{cases}}\)
đến đây dễ rồi
b, =>(x-18)(x+16)=(x+4)(x-17)
=>x2+16x-18x-288=x2-17x+4x-68
=>x2-2x-288-x2+13x+68=0
=>11x-220=0
=>11x=220
=>x=20
\(\left(x+\frac{1}{4}\right)^2+\frac{11}{25}=\frac{18}{25}\)
\(\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{7}{25}\)
\(\Rightarrow x+\frac{1}{4}=\hept{\begin{cases}\sqrt{\frac{7}{25}}\\-\sqrt{\frac{7}{25}}\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}\sqrt{\frac{7}{25}}-\frac{1}{4}\\-\sqrt{\frac{7}{25}}-\frac{1}{4}\end{cases}}\)
\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)
\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)
vậy x=-20 và y=-2
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\left(x-1\right)\cdot3=8\cdot9\)
\(\left(x-1\right)\cdot3=72\)
\(x-1=\frac{72}{3}\)
\(x-1=24\)
\(x=24+1\)
\(x=25\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(x\cdot\left(x+1\right)=18\cdot4\)
\(x\left(x+1\right)=72\)
\(x\left(x+1\right)=8\cdot9\)
\(x=8\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\frac{x}{-4}=\frac{-9}{x}\)
\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)
\(x^2=36\)
\(x^2=\left(-6\right)^2\)hoặc \(x^2=6^2\)
\(x=-6\) hoặc\(x=6\)
a)x-1/9=24/9 => x-1=24 =>x=23
b)x(x+1)=18*4 =>x=8
c)-x:4=-9:x =>-1.x2=-1.36 =>x=6
k mik nha!
ta có :\(\frac{-1}{3}< \frac{x}{36}< \frac{y}{18}< \frac{-1}{4}\)
=\(\frac{-12}{36}< \frac{x}{36}< \frac{y.2}{36}< \frac{-9}{36}\)
=\(\frac{-12}{36}< \frac{-11}{36}< \frac{-10}{36}< \frac{-9}{36}\)
nếu \(\frac{y.2}{36}=\frac{-10}{36}\)
thì : -10 : 2 = -5
=>\(\frac{-1}{3}< \frac{-11}{36}< \frac{-5}{36}< \frac{-1}{4}\)
\(\frac{x}{4}=\frac{18}{x+1}\)
=> x.(x+1)=4.18
=> x.(x+1)=72
=> x.(x+1)=8.9
=> x.(x+1)=8.(8+1)
Vậy x=8.
=> x(x+1)=72 => x= 8 hoặc x=-9