a) \(2x^2+3x-8=0\)

Ta có: \(\Delta=3^2+4.2.8=73\)

pt có 2 nghiệm

\(x_1=\frac{-3+\sqrt{73}}{4}\);\(x_1=\frac{-3-\sqrt{73}}{4}\)

d) \(\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)

Đặt \(x^2+2x=t\)

\(pt\Leftrightarrow t^2-2t-3=0\)

Ta có: \(\Delta=2^2+4.3=16,\sqrt{\Delta}=4\)

pt trên có 2 nghiệm

\(x_1=\frac{2+4}{2}=3;x_2=\frac{2-4}{2}=-1\)

\(\Rightarrow\orbr{\begin{cases}x^2+2x=3\\x^2+2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)\left(x-1\right)=0\\\left(x+1\right)^2=0\end{cases}}\)

\(\Rightarrow x\in\left\{-3;-1;1\right\}\)

c) \(x^4+8x^3+19x^2+12x=0\)

\(\Leftrightarrow x^4+4x^3+4x^3+16x^2+3x^2+12x=0\)

\(\Leftrightarrow\left(x^4+4x^3+3x^2\right)+\left(4x^3+16x^2+12x\right)=0\)

\(\Leftrightarrow x\left(x^3+4x^2+3x\right)+4\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+x^2+3x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2+3x\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow x\in\left\{0;-1;-3;-4\right\}\)

b) \(2x^4-6x^3+x^2+6x-3=0\)

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow\left(2x^4-6x^3+3x^2\right)-\left(2x^2-6x+3\right)=0\)

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

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(2x^2-6x+3\right)=0\)

\(TH1:\left(x+1\right)\left(x-1\right)=0\Leftrightarrow x=\pm1\)

\(TH2:2x^2-6x+3=0\)

Ta có: \(\Delta=6^2-4.2.3=12,\sqrt{\Delta}=\sqrt{12}\)

pt có 2 nghiệm

\(x_1=\frac{6+\sqrt{12}}{4}=\frac{3+\sqrt{3}}{2}\);\(x_1=\frac{6-\sqrt{12}}{4}=\frac{3-\sqrt{3}}{2}\)

\(-B=\left(x^2-3x\right)\left(x^2-3x+10\right)-2010=\left(x^2-3x+5\right)^2-2035\).

Ta có \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\).

Do đó \(-B\ge\left(\dfrac{11}{4}\right)^2-2035=\dfrac{-32439}{16}\Rightarrow B\le\dfrac{32439}{16}\).

...

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

\(\Leftrightarrow6x^2-4x+9x-6=0\)

\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

b)\(6x^2-13x+6=0\)

\(\Leftrightarrow6x^2-4x-9x+6=0\)

\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

c)\(10x^2-13x-3=0\)

\(\Leftrightarrow10x^2-15x+2x-3=0\)

\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)

d)\(20x^2+19x-3=0\)

\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)

\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)

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

\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)

Suy ra vô nghiệm

\(x^3-6x^2-19x+84=0\)

\(\Leftrightarrow\left(x^3-3x^2\right)-\left(3x^2-9x\right)-\left(28x-84\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-28\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-28\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2-3x-28=0\end{cases}}\)

Ta có :  \(x^2-3x-28=0\)

\(\Leftrightarrow\left(x^2-7x\right)+\left(4x-28\right)=0\)

\(\Leftrightarrow x\left(x-7\right)+4\left(x-7\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{3;-4;7\right\}\)

\(6x^2-x-2=6x^2+3x-4x-2=3x\left(2x+1\right)-2\left(2x+1\right)=\left(2x+1\right)\left(3x-2\right)\)

\(12x^2+19x+4=12x^2+3x+16x+4=3x\left(4x+1\right)+4\left(4x+1\right)=\left(3x+4\right)\left(4x+1\right)\)

\(-6x^2+11x+4=\) / câu này thấy hơi kì :> bạn xem lại đề đc hog?

\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(2x+3\right)\left(4x^2-6x+9\right)=19x\left(x^2-1\right)\)

\(\Leftrightarrow27x^3-8-8x^3-27=19x^3-19x\)

\(\Leftrightarrow19x^3-35=19x^3-19x\)

\(\Leftrightarrow35=19x\)

\(\Leftrightarrow x=\frac{35}{19}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{35}{19}\right\}\)

a: =>|x-3|=4-x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(4-x-x+3\right)\left(4-x+x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(7-2x\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{7}{2}\)

b: =>|x-5|=3-19x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(x-5-3+19x\right)\left(x-5+3-19x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(20x-8\right)\left(-18x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{1}{9}\right\}\)

c: =>\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

=>căn x-3=0

=>x=3

Đặt phép chia ta được kết quả : \(6x^3-19x^2-2=\left(3x^2-5x+1\right)\left(2x-3\right)+\left(1-17x\right)\)

Để phép chia hết => 1 -17x = 0 => x = 1/17