x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

tl

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

^HT^

Đáp án B

a: =>3x^2-3x-2x+2=0

=>(x-1)(3x-2)=0

=>x=2/3 hoặc x=1

b: =>2x^2=11

=>x^2=11/2

=>\(x=\pm\dfrac{\sqrt{22}}{2}\)

c: Δ=5^2-4*1*7=25-28=-3<0

=>PTVN

f: =>6x^4-6x^2-x^2+1=0

=>(x^2-1)(6x^2-1)=0

=>x^2=1 hoặc x^2=1/6

=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)

d: =>(5-2x)(5+2x)=0

=>x=5/2 hoặc x=-5/2

e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2

=>3x^2+5x-8=0 và x>=-1/2

=>3x^2+8x-3x-8=0 và x>=-1/2

=>(3x+8)(x-1)=0 và x>=-1/2

=>x=1

\(PT\Leftrightarrow x^2-2x+\sqrt{6x^2-12x+7}=0\\ \Leftrightarrow x^2-2x+1+\sqrt{6x^2-12x+7}-1=0\\ \Leftrightarrow\left(x-1\right)^2+\dfrac{6\left(x-1\right)^2}{\sqrt{6x^2-12x+7}+1}=0\\ \Leftrightarrow\left(x-1\right)\left(x-1+\dfrac{6}{\sqrt{6x^2-12x+7}+1}\right)=0\\ \Leftrightarrow x=1\left(x-1+\dfrac{6}{\sqrt{6x^2-12x+7}+1}>0\right)\)

em cảm ơn 

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

a. Phương trình tương đương với \(\left(x^2-2x-2\right)\left(x^2+5x-2\right)=0\)  hay \(x^2-2x-2=0\)  hoặc \(x^2+5x-2=0\). Đến đây sử dụng Delta hoặc viết hai phương trình dưới dạng \(\left(x-1\right)^2=3,\left(2x+5\right)^2=33\) ta được bốn nghiệm là \(x=1\pm\sqrt{3},-\frac{5}{2}\pm\frac{\sqrt{33}}{2}\)

b. Phương trình tương đương với \(3\left(x+5\right)\left(x+6\right)\left(x+9\right)=8x+6\left(x+5\right)\left(x+6\right)\leftrightarrow3\left(x+5\right)\left(x+6\right)\left(x+9\right)=\left(x+9\right)\left(6x+20\right)\)

hay \(\left(x+9\right)\left(3x^2+27x+70\right)=0\leftrightarrow x=-9.\)

ĐKXĐ: \(x\in R\)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

=>\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x-4=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x+1-5=0\)

=>\(\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+14}-3+\left(x+1\right)^2=0\)

=>\(\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+14-9}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>

\(\dfrac{3x^2+6x+3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+5}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\dfrac{3\left(x^2+2x+1\right)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x^2+2x+1\right)}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+14}+3}+1\right)=0\)

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

=>x+1=0

=>x=-1(nhận)