De phuong trinh co 2 nghiem thi \(\Delta\ge0\)

\(4^2-4\left(m^2-m+1\right)\ge0\)

\(< =>16-4m^2+4m-4\ge0\)

\(< =>4m^2-4m-12\le0\)

\(< =>\left(2m-1\right)^2\le13\)

\(< =>-\sqrt{13}\le2m-1\le\sqrt{13}\)

\(< =>\dfrac{1-\sqrt{13}}{2}\le m\le\dfrac{\sqrt{13}+1}{2}\)

DKXD : \(x\ge-1;y\ne-1\)

Dat : \(\left\{{}\begin{matrix}\sqrt{x+1}=a\left(a\ge0\right)\\y+1=b\left(b\ne0\right)\end{matrix}\right.\)

hpt<=>\(\left\{{}\begin{matrix}a+2-\dfrac{2}{y+1}=2\\2a-\dfrac{1}{y+1}=\dfrac{3}{2}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}a+2-\dfrac{2}{b}=2\\2a-\dfrac{1}{b}=\dfrac{3}{2}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}a-\dfrac{2}{b}=0\\4a-\dfrac{2}{b}=3\end{matrix}\right.< =>\left\{{}\begin{matrix}3a=3\\a=\dfrac{2}{b}\end{matrix}\right.< =>\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)(tmdk)

\(=>\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)(tmdk)

a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)

\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)

\(< =>\dfrac{x-5}{4}=4x-17\)

\(< =>x-5=16x-68\)

\(< =>15x=68-5=63\)

\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)

b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)

\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)

\(< =>2x+1+1+2\sqrt{2x+1}=4x\)

\(< =>2x-2\sqrt{2x+1}-2=0\)

\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)

\(< =>\left(\sqrt{2x+1}-1\right)^2=4\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)

\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)

Theo he thuc viet thi \(\left\{{}\begin{matrix}x_1.x_2=-3\\x_1+x_2=1\end{matrix}\right.\)

\(x_1^3x_2+x_2^3x_1+21=x_1x_2\left(x_1^2+x_2^2\right)+21\)

\(=-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+21=-3\left(1+6\right)+21=0\)

a , de phuong trinh co 2 nghiem trai dau thi 

\(\left\{{}\begin{matrix}\Delta\ge0\\x_1.x_2\le0\end{matrix}\right.< =>\left\{{}\begin{matrix}4m^2-4\left(2m-1\right)\ge0\\2m-1\le0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}4m^2-8m+4m\ge0\\m\le\dfrac{1}{2}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\left(m-1\right)^2\ge0\\m\le\dfrac{1}{2}\end{matrix}\right.\)

\(< =>m\le\dfrac{1}{2}\)

b, Theo he thuc viet thi : \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1.x_2=2m-1\end{matrix}\right.\)khi do :

pt \(< =>2\left(x_1^2+2x_1x_2+x_2^2\right)-9x_1x_2=27\)

\(< =>2\left(x_1+x_2\right)^2-9x_1x_2-27=0\)

\(< =>8m^2-18m-18=0\)

\(< =>4m^2-9m-9=0\)

\(< =>\left(2m\right)^2-2.2m.\dfrac{9}{4}+\dfrac{81}{16}-9-5\dfrac{1}{6}=0\)

\(< =>\left(2m-\dfrac{9}{4}\right)^2=14\dfrac{1}{6}\) 

\(< =>\left(2m-\dfrac{9}{4}\right)^2=\dfrac{85}{6}\)

\(< =>\left[{}\begin{matrix}m=\dfrac{\sqrt{\dfrac{85}{6}}+\dfrac{9}{4}}{2}\\m=\dfrac{-\sqrt{\dfrac{85}{6}}+\dfrac{9}{4}}{2}\end{matrix}\right.\)

doi chieu dieu kien nua nhe

1) \(x^2-5x-6=0\)

\(< =>x^2+x-6x-6=0\)

\(< =>x\left(x+1\right)-6\left(x+1\right)=0\)

\(< =>\left(x-6\right)\left(x+1\right)=0\)

\(< =>\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

a, Để phương trình có 2 nghiệm phân biệt thì \(\Delta>0\)

\(< =>\left[-2\left(m-1\right)\right]^2-4\left(-2m+5\right)>0\)

\(< =>4m^2-8m+4+8m-20>0\)

\(< =>m^2-4>0< =>m^2>4< =>\left[{}\begin{matrix}m>2\\m< -2\end{matrix}\right.\)

b, Theo hệ thức Viét thì bất phương trình tương đương với :

\(2m-2+2\left(-2m+5\right)\le26\)

\(< =>2m+26-10+2\ge0\)

\(< =>2m+18\ge0\)

\(< =>m\ge-9\)

Kết hợp với điều kiện ta có \(-2>m\ge-9\)

\(B=\left(\dfrac{y+2}{y\sqrt{y}-1}+\dfrac{\sqrt{y}}{y+\sqrt{y}-1}+\dfrac{1}{1-\sqrt{y}}\right):\dfrac{\sqrt{y}-1}{2}\)

\(=\left[\dfrac{y+2}{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}+\dfrac{\sqrt{y}\left(\sqrt{y}-1\right)}{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}-\dfrac{y+\sqrt{y}+1}{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}\right]:\dfrac{\sqrt{y}-1}{2}\)

\(=\left[\dfrac{y+2+y-\sqrt{y}-y-\sqrt{y}-1}{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}\right].\dfrac{2}{\sqrt{y}-1}\)

\(=\dfrac{y-2\sqrt{y}+1}{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}.\dfrac{2}{\sqrt{y}-1}=\dfrac{2}{y+\sqrt{y}+1}\)