\(\frac{189}{209}\) x \(\frac{1291}{3290}\)= ?
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\(\frac{149-x}{26}+\frac{171-x}{24}+\frac{189-x}{22}+\frac{203-x}{20}=10\)
\(\left(\frac{149-x}{26}-1\right)+\left(\frac{171-x}{24}-2\right)+\left(\frac{189-x}{22}-3\right)+\left(\frac{203-x}{20}-4\right)=0\)
\(\frac{123-x}{26}+\frac{123-x}{24}+\frac{123-x}{22}+\frac{123-x}{20}=0\)
\(\left(123-x\right)\left(\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\right)=0\)
\(123-x=0\left(vì\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\ne0\right)\)
\(\Rightarrow x=123\)
ta có
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)\(-4\)
\(\Rightarrow\left(\frac{x+1}{212}+1\right)+\left(\frac{x+2}{211}+1\right)+\left(\frac{x+3}{210}+1\right)+\left(\frac{x+4}{209}+1\right)=-4+4\)
=> \(\frac{x+1+212}{212}+\frac{x+2+211}{211}+\frac{x+3+210}{210}+\frac{x+4+209}{209}\) =\(0\)
=> \(\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}\)=\(0\)
=> (x+213) \(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)=0
mà\(\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)\)\(\ne0\)
=>x+213=0 => x=-213
vậy x= -213
\(\frac{x+1}{212}+\frac{x+2}{211}+\frac{x+3}{210}+\frac{x+4}{209}=-4\)
\(\Rightarrow\frac{x+1}{212}+1+\frac{x+2}{211}+1+\frac{x+3}{210}+1+\frac{x+4}{209}+1=-4+4=0\)
\(\Rightarrow\frac{x+213}{212}+\frac{x+213}{211}+\frac{x+213}{210}+\frac{x+213}{209}=0\)
\(\Rightarrow\left(x+213\right)\left(\frac{1}{212}+\frac{1}{211}+\frac{1}{210}+\frac{1}{209}\right)=0\)
\(\Rightarrow x+213=0\Leftrightarrow x=-213\)
nhân cả 2 vế với x, ta có: \(\frac{x^2}{3}=-\frac{x.y}{7}\)thay x.y=-189, ta có: \(\frac{x^2}{3}=\frac{189}{7}\)=> \(x^2\).7=189.3 => x^2 = 81 => x=9 ( vì x>y nên x=9) => y=-21 => x+y=-12
Đặt \(\frac{x}{3}=k;\frac{y}{7}=k\Rightarrow x=3k;y=7k\)(1)
Mà xy=189(2)
Từ (1) và (2) suy ra:3k.7k=189
21k2=189
k2=9=32=(-3)2
\(TH1:\left[\begin{array}{nghiempt}\frac{x}{3}=3\\\frac{y}{7}=3\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\y=21\end{array}\right.\)
\(TH2:\left[\begin{array}{nghiempt}\frac{x}{3}=-3\\\frac{y}{7}=-3\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-9\\y=-21\end{array}\right.\)
Vậy TH1:x=9;y=21
TH2:x=-9;y=-21
Cách 1. Đặt \(\frac{x}{3}=\frac{y}{7}=k\) \(\Rightarrow\begin{cases}x=3k\\y=7k\end{cases}\)
\(\Rightarrow xy=189\Leftrightarrow3k.7k=189\Leftrightarrow k^2=9\Leftrightarrow k=\pm3\)
Nếu k = 3 thì x = 9 , y = 21
Nếu k = -3 thì x = -9 , y = -21
Ta có:
\(\frac{x}{3}=\frac{-y}{7}\Rightarrow x=\frac{-3y}{7}\)
Thay vào: x . y, ta được:
\(x\cdot y=\frac{-3y}{7}\cdot y=\frac{-3y^2}{7}=-189\)
=> -3y2 = -189 * 7 = -1323
=> y2 = -1323 : (-3) = 441
=> y = 21 hoặc y = -21
x . y = -189
=> x = -189 : 21 = -9 hoặc x = -189 : (-21) = 9
mà x > y
Vậy x = 9; y = -21
=> \(\left(\frac{x}{3}\right)^2=\left(\frac{x}{3}\right).\left(-\frac{y}{7}\right)=\frac{-xy}{21}=\frac{189}{21}=9\)
=> x/3 = 3 hoặc x/3 = - 3
+) x/3 = 3 => x = 9 => y = -21 < 9 => Loại
+) x/3 = - 3 => x = - 9 => y = -189: (-9) = 21 > -9 (Thỏa mãn x< y)
Vạy x = - 9
\(\frac{189}{209}\)x\(\frac{1291}{3290}=\frac{3483}{9823}\)
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