\(\frac{149-x}{26}+\frac{171-x}{24}+\frac{189-x}{22}+\frac{203-x}{20}=10\)

\(\left(\frac{149-x}{26}-1\right)+\left(\frac{171-x}{24}-2\right)+\left(\frac{189-x}{22}-3\right)+\left(\frac{203-x}{20}-4\right)=0\)

\(\frac{123-x}{26}+\frac{123-x}{24}+\frac{123-x}{22}+\frac{123-x}{20}=0\)

\(\left(123-x\right)\left(\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\right)=0\)

\(123-x=0\left(vì\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\ne0\right)\)

\(\Rightarrow x=123\)

x = -213 nha bạn 

ai trả lời câu này mk cho

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

Cách 2. Từ \(\frac{x}{3}=\frac{y}{7}\Rightarrow y=\frac{7x}{3}\) thay vào giả thiết xy = 189 được  :

\(x.\frac{7x}{3}=189\Leftrightarrow\frac{7x^2}{3}=189\Leftrightarrow x^2=81\Leftrightarrow x=\pm9\)

Với x = 9 => y = 21

Với 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

sorry thêm nữa nhé:

đặt x/3=-y/7=k

=> x=3k; y=-7k

xy=-189

=> 3k.(-7k)=-189

=> -21k²=-189

=> k²=-189:(-21)

=> k²=9

=> k²=3²=(-3)²

=> k=3 hoặc k=-3

TH1: k=3

=> x=3k=3.3=9

=> y=-7k=-7.3=-21

=> x+y=9+(-21)=-12

TH2: k=-3

=> x=3k=3.(-3)=-9

=> y=-7k=(-7).(-3)=21

mà x > y => loại TH2.

=> \(\left(\frac{x}{3}\right)^2=\left(\frac{x}{3}\right).\left(-\frac{y}{7}\right)=\frac{-xy}{21}=\frac{189}{21}=9\)

=> 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

cô loan ơi giải giúp em với