b: \(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-18\right)=-36\)

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

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

hay \(x\in\left\{0;-3;\dfrac{-3+\sqrt{73}}{2};\dfrac{-3-\sqrt{73}}{2}\right\}\)

c: \(\Leftrightarrow6x^4-18x^3-17x^3+51x^2+11x^2-33x-2x+6=0\)

\(\Rightarrow\left(x-3\right)\left(6x^3-17x^2+11x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^3-12x^2-5x^2+10x+x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^2-5x+1\right)=0\)

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

hay \(x\in\left\{3;2;\dfrac{1}{3};\dfrac{1}{2}\right\}\)

d: \(\Leftrightarrow\left(x-1\right)^2\cdot\left(x^2+3x+1\right)=0\)

hay \(x\in\left\{1;\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)

câu b nè : http://123link.pw/fGAhMX

1: ĐKXĐ: x>=-1

\(\begin{cases}\sqrt{x+1}=\sqrt{2}(8y^2+8y+1)\left(1\right)\\ 4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\left(2\right)\end{cases}\)

(2): \(4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\)

=>\(8x^3 - 64y^3 - 12x^2 - 48y^2 + 6x + 12y - 2 = 0\)

=>\((8x^3-12x^2+6x-1)-(64y^3+48y^2+12y+1)=0\)

=>\((2x-1)^3-(4y+1)^3=0\)

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

=>2x-1=4y+1

=>2x-4y=2

=>x-2y=1

=>x=2y+1

(1) sẽ tương đương: \(\sqrt{2y + 1 + 1} = \sqrt{2}(8y^2 + 8y + 1)\)

=>\(\sqrt{2(y + 1)}=\sqrt{2}(8y^2+8y+1)\)

=>\(\sqrt{y + 1}=8y^2+8y+1\)

Đặt \(t = 8y^2 + 8y + 1 \ge 0\)

Ta có: \(\sqrt{y + 1}=8y^2+8y+1\)

=>\(\sqrt{y+1}=t\)

=>\(y+1=t^2\)

Phương trình sẽ trở thành:

\(8(t^2-1)^2+8(t^2-1)+1=t\)

=>\(8t^4-8t^2-t+1=0\)

=>\((t-1)(8t^3+8t^2-1)=0\)

=>t-1=0

=>t=1

=>8y(y+1)=0

=>y=0 hoặc y=-1

Khi y=0 thì x=1(nhận)

Khi y=-1 thì x=-1(loại)

4x³ +6x² -17x-22 = 4(x-2)\(\sqrt{6x+10}\)

<=> (x-2)(4x²+14x+11)=4(x-2)\(\sqrt{6x+10}\)

<=> (x-2)(4x²+14x+11-4\(\sqrt{6x+10}\))=0

TH1:x=2

TH2:tự xét nha

.

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

1. \(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)< 0\)

\(\Rightarrow-\frac{1}{3}< x< \frac{1}{2}\)

2. \(\Leftrightarrow\left(x-2\right)\left(3-2x\right)>0\)

\(\Rightarrow\frac{3}{2}< x< 2\)

3. \(\Leftrightarrow\left(5x-3\right)^2>0\)

\(\Rightarrow x\ne\frac{3}{5}\)

4. \(\Leftrightarrow-3\left(x-\frac{1}{6}\right)-\frac{59}{12}< 0\)

\(\Rightarrow x\in R\)

5. \(\Leftrightarrow2\left(x-1\right)^2+5\ge0\)

\(\Rightarrow x\in R\)

6. \(\Leftrightarrow\left(x+2\right)\left(8x+7\right)\le0\)

\(\Rightarrow-2\le x\le-\frac{7}{8}\)

7.

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

\(\Rightarrow x\in R\)

8. \(\Leftrightarrow\left(3x-2\right)\left(2x+1\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge\frac{2}{3}\end{matrix}\right.\)

9. \(\Leftrightarrow\frac{1}{3}\left(x+3\right)\left(x+6\right)< 0\)

\(\Rightarrow-6< x< -3\)

10. \(\Leftrightarrow x^2-6x+9>0\)

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

\(\Rightarrow x\ne3\)