\(\frac{k\left(x+2\right)-3\left(k-1\right)}{x+1}=1\)

\(\Leftrightarrow\left(k-1\right)x=2-k\)

Với \(k=1\) thì phương trình vô nghiệm

Với \(k\ne1\)thì

\(x=\frac{2-k}{k-1}>0\)

\(\Leftrightarrow1< k< 2\)

\(\text{để phương trình có nghiệm duy nhất thì pt phải có nghiệm kép}\)

x>=1/2 thì: \(x^2-6x+5-k-2xk=\left(x+a\right)^2\text{ hay: }-6x-2xk+5-k=2xa+a^2\text{ do đó: }-6x-2xk=2xa;5-k=a^2\Rightarrow-3-k=2a;5-k=a^2\Rightarrow8=a^2-2a\Leftrightarrow a^2-2a-8=\left(a-4\right)\left(a+2\right)=0\text{ hay }a=4\text{ hoặc: }a=-2\Rightarrow k=1\text{ hoặc: }k=-11\text{ tương tự với TH còn lại.}\)

\(a,< =>\Delta=0\)

\(=>[-\left(k+1\right)]^2-4\left(2+k\right)=0\)

\(< =>k^2+2k+1-8-4k=0\)

\(< =>k^2-2k-7=0\)

\(\Delta1=\left(-2\right)^2-4\left(-7\right)=32>0\)

\(=>\left[{}\begin{matrix}k1=\dfrac{2+\sqrt{32}}{2}\\k2=\dfrac{2-\sqrt{32}}{2}\end{matrix}\right.\)

b,\(< =>\Delta'=0< =>\left(k-1\right)^2-\left(k+9\right)=0\)

\(< =>k^2-2k+1-k-9=0< =>k^2-3k-8=0\)

\(\Delta=\left(-3\right)^2-4\left(-8\right)=41>0\)

\(=>\left[{}\begin{matrix}k1=\dfrac{3+\sqrt{41}}{2}\\k2=\dfrac{3-\sqrt{41}}{2}\end{matrix}\right.\)

a) \(\text{Δ}=\left[-\left(k+1\right)\right]^2-4\cdot1\cdot\left(k+2\right)\)

\(=k^2+2k+1-4k-8\)

\(=k^2-2k-7\)

Để phương trình có nghiệm kép thì Δ=0

\(\Leftrightarrow k^2-2k-7=0\)(1)

\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(-7\right)=4+28=32\)

Vì Δ>0 nên phương trình (1) có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}k_1=\dfrac{2-4\sqrt{2}}{2}=1-2\sqrt{2}\\k_2=\dfrac{2+4\sqrt{2}}{2}=1+2\sqrt{2}\end{matrix}\right.\)

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Xét phương trình: \(\frac{2x}{3}+\frac{2x-1}{5}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}+\frac{2x-1}{5}=4\)

\(\Leftrightarrow x+\frac{2x-1}{5}=4\Leftrightarrow\frac{5x+2x-1}{5}=4\)

\(\Leftrightarrow7x-1=20\Leftrightarrow x=3\)

Để hai phương trình \(\frac{2x}{3}+\frac{2x-1}{5}=4-\frac{x}{3}\)và \(\left(k+1\right)x+k=26\)tương đương thì:

x = 3 là nghiệm của \(\left(k+1\right)x+k=26\)

\(\Rightarrow3\left(k+1\right)+k=26\Leftrightarrow3k+3+k=26\)

\(\Leftrightarrow4k=23\Leftrightarrow k=\frac{23}{4}\)

Vậy \(k=\frac{23}{4}\)thì hai phương trình trên tương đương

Ai làm đc câu nào thì làm giúp mình với ạ, cảm ơn trc:(((

\(1,3x-5x+5=-8\)

\(\Leftrightarrow-2x+5+8=0\)

\(\Leftrightarrow-2x=-13\)

\(\Leftrightarrow x=\frac{13}{2}\)

Hướng dẫn:

Giải pt đầu tiên => nghiệm x₀ (nghiệm ngày bạn tự tìm)

Thay vào pt sau: (k+1)x + k =26

Tức là (k+1) x₀ +k =26 . Từ đó tìm k.