Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0

=x(x+3)+2(x+3)=(x+2)(x+3)=0

Dễ rồi

2)\(x^2-x-6=0=x^2-3x+2x-6=0\)

=x(x-3)+2(x-3)=0

=(x+2)(x-3)=0

Dễ rồi

3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)

Vì \(x^2+1>0\)

=>\(\left(x+2\right)^2=0\)

Dễ rồi

4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0

=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)

=>x+1=0

=>..................

5)\(x^2-7x+6=x^2-6x-x+6\) =0

=x(x-6)-(x-6)=0

=(x-1)(x-6)=0

=>.....

6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0

=2x(x+1)-5(x+1)=0

=(2x-5)(x+1)=0

7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0

Dễ rồi

Nghỉ đã hôm sau làm mệt

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

\(\Leftrightarrow\left(x-1\right)^2>=0\)

hay \(x\in R\)

a: =>7-x=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

a: =>-x+7=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

a.

\(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2x+10-x^2-5x=0\)

\(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

b.

\(2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-5}{2}\end{matrix}\right.\)

a) 2(x + 5) - x² - 5x = 0

\(\Leftrightarrow\) 2x + 10 - x² - 5x = 0

\(\Leftrightarrow\) -x² - 3x + 10 = 0

\(\Leftrightarrow\) -(x² + 3x - 10) = 0

\(\Leftrightarrow\) x² + 3x - 10 = 0

\(\Leftrightarrow\) x² - 2x + 5x - 10 = 0

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

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy.........

b) 2x² + 3x - 5 = 0

\(\Leftrightarrow\) 2x² - 2x + 5x - 5 = 0

\(\Leftrightarrow\) 2x(x - 1) + 5(x - 1) = 0

\(\Leftrightarrow\) (2x + 5)(x - 1) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

Vậy.........

c) (x- 1)² + 4(x + 2) - (x² - 3) = 0

\(\Leftrightarrow\) x² - 2x + 1 + 4x + 8 - x² + 3 = 0

\(\Leftrightarrow\) 2x + 4 = 0

\(\Leftrightarrow\) 2(x + 2) = 0

\(\Leftrightarrow\) x = -2

Vậy.............

Mấy bn đọc bài mk xong nhớ tik nha

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)

\(B=\dfrac{4x_1-1}{x_2}+\dfrac{4x_2-1}{x_1}=\dfrac{4x_1^2-x_1+4x_2^2-x_2}{x_1x_2}\)

\(=\dfrac{4\left(x_1+x_2\right)^2-8x_1x_2-\left(x_1+x_2\right)}{x_1x_2}=\dfrac{4.\left(-\dfrac{3}{2}\right)^2-8.\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{2}\right)}{-\dfrac{1}{2}}=-29\)

a: Khi m=-2 thì phương trình trở thành \(x^2+2x-3=0\)

=>(x+3)(x-1)=0

=>x=-3 hoặc x=1

b: \(\text{Δ}=\left(-2\right)^2-4\left(m-1\right)=4-4m+4=-4m+8\)

Để phương trình có hai nghiệm phân biệt thì -4m+8>0

=>-4m>-8

hay m<2

Theo hệ thức Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=m-1\end{matrix}\right.\)

Theo đề, ta có: 

\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-2x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=-\dfrac{2}{3}\\x_1=2x_2=-\dfrac{4}{3}\end{matrix}\right.\)

Ta có: \(x_1x_2=m-1\)

\(\Leftrightarrow m-1=\dfrac{8}{9}\)

hay m=17/9(nhận)

a. Thay m=-2 ta được: \(x^2+2x-2-1=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

b. Để phương trình (*) có 2 nghiệm phân biệt thì \(\Delta=4-4\left(m-1\right)>0\Leftrightarrow1>m-1\Leftrightarrow m< 2\)

Áp dụng định lí Vi-et ta có: \(x_1+x_2=\dfrac{-2}{1}=-2\)

Ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-2x_2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-\dfrac{4}{3}\\x_2=\dfrac{-2}{3}\end{matrix}\right.\)\(\Rightarrow x_1.x_2=\dfrac{m-1}{1}=\dfrac{-4}{3}.\dfrac{-2}{3}=m-1\Rightarrow m=\dfrac{17}{9}\)<2

Vậy m=\(\dfrac{17}{9}\)

a, Khi m=-2 thay vào pt ta đc:

x²+2x-2-1=0  =>  x²+2x-3=0 có a=1, b=2 -> b'=1, c=-3

△'=b'²-ac=1-1.(-3)=4

△'>0 nên pt có 2n_o pb:

\(x_1=\dfrac{-b'^{^2}+\sqrt{\Delta'}}{a}=1\); \(x_2=-3\)