Lmj thấy phương trình nào??

b, \(\Delta'=b'^2-ac=\left[-\left(m-1\right)\right]^2-1.\left(-m-3\right)=m^2-2m+1+m+3\)

\(=m^2-m+4=m^2-m+\frac{1}{4}+\frac{15}{4}=\left(m-\frac{1}{2}\right)^2+\frac{15}{4}>0\)

Vậy pt (1) có 2 nghiệm x1,x2 với mọi m

Theo hệ thức vi-et ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m-1\right)\left(2\right)\\x_1x_2=-m-3\left(3\right)\end{cases}}\)

Ta có: \(x_1^2+x_2^2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\)

<=>\(4\left(m-1\right)^2-2\left(-m-3\right)=10\)

<=>\(4m^2-8m+4+2m+6=10\)

<=>\(4m^2-6m+10=10\Leftrightarrow2m\left(2m-3\right)=0\)

<=>\(\orbr{\begin{cases}m=0\\m=\frac{3}{2}\end{cases}}\)

c, Từ (2) => \(m=\frac{x_1+x_2+2}{2}\)

Thay m vào (3) ta có: \(x_1x_2=\frac{-x_1-x_2-2}{2}-3=\frac{-x_1-x_2-8}{2}\)

<=>\(2x_1x_2+x_1+x_2=-8\)

22.

ĐKXĐ: \(y\ne1\)

\(\left\{{}\begin{matrix}x^2-\dfrac{1}{y-1}=2\\2x^2+\dfrac{3}{1-y}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+\dfrac{2}{1-y}=4\\2x^2+\dfrac{3}{1-y}=2\end{matrix}\right.\)

Trừ pt dưới cho trên:

\(\Rightarrow\dfrac{1}{1-y}=-2\)

\(\Rightarrow1-y=-\dfrac{1}{2}\Rightarrow y=\dfrac{3}{2}\)

Thế vào \(x^2-\dfrac{1}{y-1}=2\)

\(\Rightarrow x^2=4\Rightarrow x=\pm2\)

Vậy nghiệm của hệ là \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right);\left(-2;\dfrac{3}{2}\right)\)

b.

ĐKXĐ: \(x\ne-\dfrac{1}{2}\)

\(Hệ\Leftrightarrow\left\{{}\begin{matrix}2y^2-\dfrac{10}{2x+1}=8\\2y^2-\dfrac{11}{2x+1}=7\end{matrix}\right.\)

Trừ pt trên cho dưới:

\(\Rightarrow\dfrac{1}{2x+1}=1\)

\(\Rightarrow2x+1=1\)

\(\Rightarrow x=0\)

Thế vào \(y^2-\dfrac{5}{2x+1}=4\)

\(\Rightarrow y^2=9\Rightarrow y=\pm3\)

Vậy nghiệm của hệ là \(\left(x;y\right)=\left(0;3\right);\left(0;-3\right)\)

\(\left\{{}\begin{matrix}x+y=5\\\dfrac{3}{5}+\dfrac{2}{x-y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\\dfrac{2}{x-y}=\dfrac{12}{5}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+y=5\\6\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\x-y=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{35}{6}\\y=x-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{35}{12}\\y=\dfrac{25}{12}\end{matrix}\right.\)

\(ĐK:x\ge1\\ PT\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\sqrt{x+2}=0\\ \Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\sqrt{x+2}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ \Leftrightarrow x\in\varnothing\)