a, <=> (x-1)^3 + x^2(x-1)=0

<=> (x-1)(x^2-2x+1+x^2)=0

<=> (x-1)(2x^2-2x+1)=0

=> x=1

2x^2-2x+1=0 (*)

giải (*):

2x^2-2x+1=0

<=> (x-1)^2 + x^2 > 0

=> * vô nghiệm

=> Pt có nghiệm là 1.

b, x^2+x-12=0

<=> (x-3)(x+4)=0

=> x=3 hoặc x = -4

vậy....

c, 6x^2-11x-10=0

<=> (x-5/2)(6x+4)=0

=> x=5/2 hoặc x= -2/3.

vậy...

a/ Đặt \(6x+7=a\Rightarrow\left\{{}\begin{matrix}6x+8=a+1\\6x+6=a-1\end{matrix}\right.\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)a^2-72=0\)

\(\Leftrightarrow\left(a^2-1\right)a^2-72=0\)

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)

\(\Leftrightarrow a^2=9\) (do \(a^2+8>0\))

\(\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne-4;-5;-6;-7\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

a) Ta có: \(2x^3+5x^2-3x=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!

a.

$x^2-11=0$

$\Leftrightarrow x^2=11$

$\Leftrightarrow x=\pm \sqrt{11}$

b. $x^2-12x+52=0$

$\Leftrightarrow (x^2-12x+36)+16=0$

$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)

Vậy pt vô nghiệm.

c.

$x^2-3x-28=0$

$\Leftrightarrow x^2+4x-7x-28=0$

$\Leftrightarrow x(x+4)-7(x+4)=0$

$\Leftrightarrow (x+4)(x-7)=0$

$\Leftrightarrow x+4=0$ hoặc $x-7=0$

$\Leftrightarrow x=-4$ hoặc $x=7$

d.

$x^2-11x+38=0$

$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$

$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)

Vậy pt vô nghiệm

e.

$6x^2+71x+175=0$

$\Leftrightarrow 6x^2+21x+50x+175=0$

$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$

$\Leftrightarrow (3x+25)(2x+7)=0$

$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$

$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$

f.

$x^2-(\sqrt{2}+\sqrt{8})x+4=0$

$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$

$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$

$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$

$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$

$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$

g.

$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$

$\Leftrightarrow (1+\sqrt{3})x^2-(1+\sqrt{3})x-(\sqrt{3}x-\sqrt{3})=0$

$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$

$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$

$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1