1-1-9-3-4444=-4456

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1-1-9-3-4444 = - 4456

Tìm x,y hả bn??

phương ly: ừ giúp mình đi

ĐKXĐ....

\(\Leftrightarrow4x^2-5x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x^3-3x+x^2-2x+1=-2\sqrt{x-1}\)

\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)^2=-2\sqrt{x-1}\)

Đặt \(\sqrt{x-1}=t\left(t\ge0\right)\Rightarrow t^2=x-1\Leftrightarrow x=t^2+1\)

\(\Leftrightarrow3\left(t^2+1\right)t^2+t^4=-2t\)

\(\Leftrightarrow3t^4+3t^2+t^4+2t=0\)

\(\Leftrightarrow4t^4+3t^2+2t=0\)

\(\Leftrightarrow t\left(4t^3+3t+2\right)=0\Leftrightarrow\orbr{\begin{cases}t=0\\t=-\frac{1}{2}\left(l\right)\end{cases}}\)

\(\Rightarrow t=0\Leftrightarrow x=1\left(tm\right)\)

Vậy x =1 

Ta có : \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

<=> \(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)

<=>  \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)

Nếu a + b + c + d = 0

=> a + b = -(c + d) 

b + c = -(a + d) 

c + d = -(a + b) 

d + a = -(b + c)

Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Nếu a + b + c + d \(\ne\)0 

=> \(\frac{1}{b+c+d}=\frac{1}{a+c+d}=\frac{1}{a+b+d}=\frac{1}{a+b+c}\)

=> b + c + d = a + c + d = a + b + d = a + b + c

=> a = b = c = d

Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{a+b}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

Vậy nếu a + b + c + d \(\ne\)0 => M = 4

nếu a + b + c + d = 0 => M = -4

CÓ MÌNH NÈ

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