Question

[Hào]

Giúp mình câu 2 phần 2 và câu 3 phần 2 với

Answer 1

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câu 2 phần 2:

\(\left\{{}\begin{matrix}4x+3y=11\\4x-y=7\end{matrix}\right.\)\(< =>\left\{{}\begin{matrix}4y=4\\4x-y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\).Vậy hệ pt có nghiệm

(x,y)=(2;1)

caau3 phần 2:

\(x^2-2x+m-1=0\)(1)

\(\Delta'=\left(-1\right)^2-\left(m-1\right)=1-m+1=2-m\)

để pt (1) có 2 nghiệm x1,x2<=>\(\Delta'\ge0< =>2-m\ge0< =>m\le2\)

theo vi ét=>\(\left\{{}\begin{matrix}x1+x2=2\left(1\right)\\x1.x2=m-1\left(3\right)\end{matrix}\right.\)

có: \(x1^4\)\(-x1^3=x2^4-x2^3\)

\(< =>x1^4-x2^4-x1^3+x2^3=0\)

\(< =>\left(x1^2-x2^2\right)\left(x1^2+x2^2\right)-\left(x1^3-x2^3\right)\)\(=0\)

\(< =>\left(x1-x2\right)\left(x1+x2\right)\left[\left(x1+x2\right)^2-2x1x2\right]\)\(-\left(x1-x2\right)\left(x1^2+x1x2+x^2\right)=0\)

\(< =>\)\(\left(x1-x2\right)\left[2.2^2-2\left(m-1\right)-\left(x1^2+x1x2+x2^2\right)\right]=0\)

\(< =>.\left(x1-x2\right)\left[8-2m+2-\left(x1+x2\right)^2+x1x2\right]=0\)

<=>\(\left(x1-x2\right)\left[10-2m-4+m-1\right]=0\)

\(< =>\left(x1-x2\right)\left(5-m\right)=0\)

\(=>\left[{}\begin{matrix}x1-x2=0\\5-m=0\end{matrix}\right.< =>\left[{}\begin{matrix}x1=x2\left(2\right)\\m=5\left(loai\right)\end{matrix}\right.\)

thế(2) vào(1)=>\(x1=x2=1\left(4\right)\)

thế (4) vào (3)=>\(m-1=1=>m=2\left(TM\right)\)

vậy m=2 thì....