Bài 1:
u-v=10
=>u=v+10
uv=11
=>\(v\left(v+10\right)=11\)
=>\(v^2+10v-11=0\)
=>(v+11)(v-1)=0
=>\(\left[{}\begin{matrix}v+11=0\\v-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}v=-11\\v=1\end{matrix}\right.\)
Khi v=-11 thì u=v+10=-11+10=-1
=>|u+v|=|-1+(-11)|=12
Khi v=1 thì u=v+10=1+10=11
=>|u+v|=|11+1|=12
Bài 2:
a: \(x-y=\dfrac{1}{4}-\sqrt{7}\)
=>\(x=y+\dfrac{1}{4}-\sqrt{7}\)
\(xy=\dfrac{\sqrt{7}}{4}\)
=>\(y\left(y+\dfrac{1}{4}-\sqrt{7}\right)=\dfrac{\sqrt{7}}{4}\)
=>\(y^2+y\left(\dfrac{1}{4}-\sqrt{7}\right)-\dfrac{\sqrt{7}}{4}=0\)(1)
\(\text{Δ}=\left(\dfrac{1}{4}-\sqrt{7}\right)^2-4\cdot\dfrac{-\sqrt{7}}{4}\)
\(=\dfrac{1}{16}-2\cdot\dfrac{1}{4}\cdot\sqrt{7}+7+\sqrt{7}\)
\(=\dfrac{113}{16}-\dfrac{1}{2}\sqrt{7}=\dfrac{113-8\sqrt{7}}{16}=\dfrac{112-2\cdot4\sqrt{7}\cdot1+1}{16}\)
\(=\dfrac{\left(4\sqrt{7}-1\right)^2}{16}\)
=>Phương trình (1) có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}y=\dfrac{-\left(\dfrac{1}{4}-\sqrt{7}\right)-\sqrt{\dfrac{\left(4\sqrt{7}-1\right)^2}{16}}}{2\cdot1}\\y=\dfrac{-\left(\dfrac{1}{4}-\sqrt{7}\right)+\sqrt{\dfrac{\left(4\sqrt{7}-1\right)^2}{16}}}{2\cdot1}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}y=\dfrac{-\dfrac{1}{4}-\sqrt{7}-\dfrac{1}{4}\left(4\sqrt{7}-1\right)}{2}=\dfrac{-\dfrac{1}{4}-\sqrt{7}-\sqrt{7}+\dfrac{1}{4}}{2}=-\sqrt{7}\\y=\dfrac{-\dfrac{1}{4}-\sqrt{7}+\sqrt{7}-\dfrac{1}{4}}{2}=\dfrac{-1}{4}\end{matrix}\right.\)
khi \(y=-\sqrt{7}\) thì \(x=\dfrac{\sqrt{7}}{4}:\left(-\sqrt{7}\right)=-\dfrac{1}{4}\)
Khi y=-1/4 thì \(y=\dfrac{\sqrt{7}}{4}:\dfrac{-1}{4}=-\sqrt{7}\)
b: \(x+y=\dfrac{1}{6};xy=-\dfrac{1}{6}\)
=>x,y là các nghiệm của phương trình:
\(A^2-\dfrac{1}{6}A-\dfrac{1}{6}=0\)
=>\(6A^2-A-1=0\)
=>\(6A^2-3A+2A-1=0\)
=>(2A-1)(3A+1)=0
=>\(\left[{}\begin{matrix}2A-1=0\\3A+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{1}{2}\\A=-\dfrac{1}{3}\end{matrix}\right.\)
vậy: \(\left(x;y\right)\in\left\{\left(\dfrac{1}{2};-\dfrac{1}{3}\right);\left(-\dfrac{1}{3};\dfrac{1}{2}\right)\right\}\)
Bài 3:
a: \(2x^2+2\left(m+1\right)x-3=0\)
a=2; b=2(m+1); c=-3
Vì \(a\cdot c=2\cdot\left(-3\right)=-6< 0\)
nên phương trình \(2x^2+2\left(m+1\right)x-3=0\) luôn có nghiệm với mọi x
