a: 2k^2+kx-10=0

Khi x=2 thì ta sẽ có: 2k^2+2k-10=0

=>k^2+k-5=0

=>\(k=\dfrac{-1\pm\sqrt{21}}{2}\)

b: Khi x=-2 thì ta sẽ có:

\(\left(-2k-5\right)\cdot4-\left(k-2\right)\cdot\left(-2\right)+2k=0\)

=>-8k-20+2k-4+2k=0

=>-4k-24=0

=>k=-6

c: Theo đề, ta có:

9k-3k-72=0

=>6k=72

=>k=12

Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3

b1           \(\frac{x+a}{x+1}+\frac{x-2}{x}=2\)

ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

\(\Leftrightarrow x\left(x+a\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow x^2+ax+x^2-x-2=2x^2+2x\)

\(\Leftrightarrow ax-3x=2\)

\(\Leftrightarrow\left(a-3\right)x=2\)

để pt vô nghiệm  thì a-3=0 <=>a=3 thì pt vô nghiệm

2,\(4x-k+4=kx+k\)

\(\Leftrightarrow4x-kx=2k-4\)

\(\Leftrightarrow\left(4-k\right)x=2k-4\)

để pt có nghiệm duy nhất thì 4-k khác 0 <=> k khác 4 thì pt có nghiệm duy nhất là\(\frac{2k-4}{4-k}\)

pt vô nghiệm thì 4-k=0 <=.>k=4 

a) \(\left(x^2-2\right)\left(k-1\right)x+2k-5=0\)

\(\Delta=\left(k-1\right)^2-2k+5\)

\(=k^2-4x+6=\left(k-2\right)^2+2>0\)

=> PT luôn có nghiệm với mọi k

x²+2(m-1)x+m²+1=0 (*) Để phương trình (*) có 2 nghiệm phân biệt khi: \(\Delta>0\) hay \(\Delta=4\left(m-1\right)^2-4\left(m^2+1\right)>0\Leftrightarrow-8m>0\Leftrightarrow m<0\left(I\right)\)

Theo giả thiết giả sử ta có: \(x_1>1,x_2<1\Rightarrow\left(x_1-1\right)\left(x_2-1\right)<0\Leftrightarrow x_1x_2-\left(x_1+x_2\right)+1<0\left(II\right)\) 

Theo Vi-et ta có: \(x_1x_2=m^2+1;x_1+x_2=-2\left(m-1\right)\) Thay vào (II) Ta có: \(m^2+1+2\left(m-1\right)+1<0\Leftrightarrow m\left(m+2\right)<0\)
Hay -2<m<0 Thỏa mãn cả (I).
Vậy -2<m<0 Thì phương trình (*) thỏa mãn điều kiện bài ra

áp dụng là ra ngay

bài này bạn áp dụng in vét nha dúng thì k