Bài 5: Công thức nghiệm thu gọn

1, \(\Delta=\left(-11\right)^2-4.1.38=121-152=-31< 0\)

\(\Rightarrow\) pt vô nghiệm

2, \(\Delta=71^2-4.6.175=5041-4200=841\)

\(x_1=\dfrac{-71+\sqrt{841}}{2.6}=\dfrac{-71+29}{12}=\dfrac{-42}{12}=-\dfrac{7}{2}\)

\(x_2=\dfrac{-71-\sqrt{841}}{2.6}=\dfrac{-71-29}{12}=\dfrac{-10}{12}=-\dfrac{25}{3}\)

3, \(\Delta=\left(-3\right)^2-5.27=9-135=-126< 0\)

⇒ pt vô nghiệm

4, \(\Delta=15^2-\left(-30\right)\left(-7,5\right)=225-225=0\)

\(\Rightarrow x_1=x_2=\dfrac{-30}{2.\left(-30\right)}=\dfrac{1}{2}\)

5, \(\Delta'=\left(-8\right)^2-4.17=64-68=-4\)

⇒ pt vô nghiệm

6, \(\Delta=4^2-4.1.\left(-12\right)=16+48=64\)

\(x_1=\dfrac{-4+\sqrt{64}}{2.1}=\dfrac{-4+8}{2}=\dfrac{4}{2}=2\)

​\(x_2=\dfrac{-4-\sqrt{64}}{2.1}=\dfrac{-4-8}{2}=\dfrac{-12}{2}=-6\)​

a; Thay x=1/4 vào A, ta được:

\(A=\left(\dfrac{1}{2}+3\right):\left(\dfrac{1}{2}-2\right)=\dfrac{7}{2}:\dfrac{-3}{2}=\dfrac{-7}{3}\)

b: \(B=\dfrac{x-2\sqrt{x}+2\sqrt{x}+2-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(\left\{{}\begin{matrix}x+y=-10\\x-y=\dfrac{288}{41}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-10\\2x=\dfrac{-122}{41}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{-61}{41}+y=-10\\x=\dfrac{-61}{41}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-349}{41}\\x=\dfrac{-61}{41}\end{matrix}\right.\)

a: Thay x=25 vào A, ta được:

\(A=\dfrac{25+5+1}{5-4}=31\)

b: \(B=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)

Thay \(x=16\)

Ta có: căn A=16 +2/căn 16=4+2/4=6/4=3/2 =)

a, \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{\sqrt{16}+2}{\sqrt{16}}=\dfrac{4+2}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)

b, \(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(\Rightarrow B=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

c,\(AB=\dfrac{\sqrt{x}+2}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(\Rightarrow AB=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(\Rightarrow AB=\dfrac{\sqrt{x}-2+4}{\sqrt{x}-2}\)

\(\Rightarrow AB=1+\dfrac{4}{\sqrt{x}-2}\)

Để AB nguyên thì \(\dfrac{4}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)

Ta có bảng:
 

Vậy \(x\in\left\{1;9;16;36\right\}\)

a, B=xx−4+1√x−2+1√x+2B=xx−4+1x−2+1x+2

⇒B=x+√x+2+√x−2(√x−2)(√x+2)⇒B=x+x+2+x−2(x−2)(x+2)

⇒B=√x(√x+2)(√x−2)(√x+2)⇒B=x(x+2)(x−2)(x+2)

AB=√x+2√x.√x√x−2AB=x+2x.xx−2

⇒AB=√x−2+4√x−2⇒AB=x−2+4x−2

4√x−2∈Z⇒√x−2∈Ư(4)4x−2∈Z⇒x−2∈Ư(4)

Ta có bảng:
 

Vậy 

\(=\left(b^2+c^2+2bc-a^2\right)\left(b^2+c^2-2bc-a^2\right)\)

\(=\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)

Với \(m=0\)

\(PT\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)

Với \(m\ne0\)

\(\Delta'=\left(m-1\right)^2-m\left(m-3\right)=m+1\)

PT vô nghiệm \(\Leftrightarrow m+1< 0\Leftrightarrow m< -1\)

PT có nghiệm kép \(\Leftrightarrow m+1=0\Leftrightarrow m=-1\)

\(\Leftrightarrow x=-\dfrac{b'}{a}=\dfrac{m-1}{2m}\)

PT có 2 nghiệm phân biệt \(\Leftrightarrow m+1>0\Leftrightarrow m>-1;m\ne0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{m-1+\sqrt{m+1}}{m}\\x=\dfrac{m-1-\sqrt{m+1}}{m}\end{matrix}\right.\)

4.

Phương trình đã cho có 2 nghiệm pb khi:

\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)>0\)

\(\Leftrightarrow2m-1>0\)

\(\Leftrightarrow m>\dfrac{1}{2}\)

5.

Pt đã cho vô nghiệm khi:

\(\Delta'=9-3\left(3k+1\right)< 0\)

\(\Leftrightarrow9k>6\)

\(\Rightarrow k>\dfrac{2}{3}\)

ĐKXĐ \(x\ne\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2x-1}+\dfrac{2x\left(2x-1\right)}{2x-1}=0\)

\(\Rightarrow1+2x\left(2x-1\right)=0\)

\(\Leftrightarrow4x^2-2x+1=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(2x-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0;\forall x\)

Pt đã cho vô nghiệm