Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\Leftrightarrow4-4\left(m-1\right)\ge0\)\(\Leftrightarrow2\ge m\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(1\right)\\x_1x_2=m-1\end{matrix}\right.\) 

\(x_1^4-x_1^3=x_2^4-x_2^3\)

\(\Leftrightarrow\left(x_1^4-x_2^4\right)-\left(x_1^3-x_2^3\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left(x_1^2+x_2^2\right)-\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left[2\left(x_1^2+x_2^2\right)-x_1^2-x_1x_2-x_2^2\right]=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)=0\)

\(\Leftrightarrow x_1-x_2=0\) (2) ( vì \(x_1^2-x_1x_2+x_2^2>0;\forall x,y\))

Từ (1) (2) \(\Rightarrow x_1=x_2=1\)

\(\Rightarrow x_1x_2=m-1=1\) \(\Leftrightarrow m=2\) (Thỏa)

Vậy...

a: Khi m=4 thì phương trình trở thành \(x^2-4x+3=0\)

=>(x-3)*(x-1)=0

=>x=3 hoặc x=1

b: \(x_1+x_2=m\)

\(x_1x_2=m-1\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=m^2-2\left(m-1\right)=m^2-2m+2\)

\(x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2\)

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

\(=m^4+4m^2+4-4m^3+4m^2-8m-2m^2+4m-2\)

\(=m^4-4m^3+2m^2-4m+2\)

x1+x2=3; x1*x2=-7

B=(x1+x2)^2-2x1x2

=9-2*(-7)=23

D=(x1+x2)^3-3x1x2(x1+x2)

=3^3-3*(-7)*3

=27+63=90

F=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=10*(-7)+69

=-1

\(C=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{3^2-4\cdot\left(-7\right)}=\sqrt{37}\)

mong bạn có thể giải thích chi tiết hơn

a: Thay \(x=\sqrt2\) vào phương trình, ta được:

\(\left(m-4\right)\cdot\left(\sqrt2\right)^2-2m\cdot\sqrt2+m+2=0\)

=>2(m-4)+m+2-2m\(\sqrt2\) =0

=>\(2m-8+m-2m\sqrt2+2=0\)

=>\(m\left(3-2\sqrt2\right)=6\)

=>\(m=\frac{6}{3-2\sqrt2}=6\left(3+2\sqrt2\right)=18+12\sqrt2\)

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac{2m}{m-4}=\frac{2\left(18+12\sqrt2\right)}{18+12\sqrt2-4}=\frac{12\left(3+2\sqrt2\right)}{12+12\sqrt2}=\frac{3+2\sqrt2}{\sqrt2+1}=\sqrt2+1\)

=>\(x_2=\sqrt2+1-\sqrt2=1\)

b: TH1: m=4

Phương trình sẽ trở thành:

\(\left(4-4\right)x^2-2\cdot4\cdot x+4+2=0\)

=>-8x+6=0

=>-8x=-6

=>x=3/4

=>Nhận

TH2: m<>4

\(\Delta=\left(-2m\right)^2-4\left(m-4\right)\left(m+2\right)\)

\(=4m^2-4\left(m^2-2m-8\right)=4m^2-4m^2+8m+32=8m+32\)

Để phương trình có nghiệm thì Δ>=0

=>8m+32>=0

=>m+4>=0

=>m>=-4

=>m>=-4 và m<>4

c: Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac{2m}{m-4};x_1x_2=\frac{c}{a}=\frac{m+2}{m-4}\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=\frac{4m^2}{\left(m-4\right)^2}-2\cdot\frac{m+2}{m-4}=\frac{4m^2-2\left(m+2\right)\left(m-4\right)}{\left(m-4\right)^2}\)

\(=\frac{4m^2-2\left(m^2-2m-8\right)}{\left(m-4\right)^2}=\frac{2m^2+4m+16}{\left(m-4\right)^2}\)

e: \(\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}\)

\(=\frac{2m}{m-4}:\frac{m+2}{m-4}=\frac{2m}{m+2}\)

\(\frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\frac{2m^2+4m+16}{\left(m-4\right)^2}:\frac{\left(m+2\right)^2}{\left(m-4\right)^2}=\frac{2m^2+4m+16}{\left(m+2\right)^2}\)

Chọn D

a: =49+25

=74

\(1234 + 678 + 766 + 1322=(1234+766)+(678+1322)=2000+2000=4000\)

\(427\times42+427\times35+427\times23=427\times\left(42+35+23\right)=427\times100=42700\)

\(248\times125-248\times24-248=248\times\left(125-24-1\right)=248\times100=24800\)