a) Ta có: \(x^2+3x-10=0\)

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

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

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

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

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

a)

\(3x^2+12x-66=0\)

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

\(\Leftrightarrow x^2+4x+4=26\Leftrightarrow (x+2)^2=26\)

\(\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\)

b)

\(9x^2-30x+225=0\)

\(\Leftrightarrow (3x)^2-2.3x.5+25+200=0\)

\(\Leftrightarrow (3x-5)^2=-200< 0\) (vô lý nên pt vô nghiệm)

c)

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

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

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

\(\Rightarrow x=-5\) hoặc $x=2$

d)

$3x^2-7x+1=0$

$\Leftrightarrow 3(x^2-\frac{7}{3}x)+1=0$

$\Leftrightarrow 3(x^2-\frac{7}{3}x+\frac{7^2}{6^2})=\frac{37}{12}$

$\Leftrightarrow 3(x-\frac{7}{6})^2=\frac{37}{12}$
$\Leftrightarrow (x-\frac{7}{6})^2=\frac{37}{36}$

$\Rightarrow x-\frac{7}{6}=\frac{\pm \sqrt{37}}{6}$

$\Rightarrow x=\frac{7\pm \sqrt{37}}{6}$

e)

$3x^2+7x+2=0$

$\Leftrightarrow 3(x^2+\frac{7}{3}x+\frac{7^2}{6^2})=\frac{25}{12}$

$\Leftrightarrow 3(x+\frac{7}{6})^2=\frac{25}{12}$

$\Leftrightarrow (x+\frac{7}{6})^2=\frac{25}{36}$

$\Rightarrow x+\frac{7}{6}=\pm \frac{5}{6}$

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

f)

$4x^2-12x+9=0$

$\Leftrightarrow (2x)^2-2.2x.3+3^2=0$

$\Leftrightarrow (2x-3)^2=0\Rightarrow 2x-3=0\Rightarrow x=\frac{3}{2}$

g) Trùng câu e

h)

$x^2-4x+1=0$

$\Leftrightarrow x^2-4x+4-3=0$

$\Leftrightarrow (x-2)^2=3\Rightarrow x-2=\pm \sqrt{3}$

$\Rightarrow x=2\pm \sqrt{3}$

i)

$2x^2-6x+1=0$

$\Leftrightarrow 2(x^2-3x+\frac{3^2}{2^2})=\frac{7}{2}$

$\Leftrightarrow 2(x-\frac{3}{2})^2=\frac{7}{2}$

$\Leftrightarrow (x-\frac{3}{2})^2=\frac{7}{4}$

$\Rightarrow x-\frac{3}{2}=\pm \frac{\sqrt{7}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{7}}{2}$

j)

$3x^2+4x-4=0$

$\Leftrightarrow 3x^2+6x-2x-4=0$

$\Leftrightarrow 3x(x+2)-2(x+2)=0$

$\Leftrightarrow (x+2)(3x-2)=0$

$\Rightarrow x+2=0$ hoặc $3x-2=0$

$\Rightarrow x=-2$ hoặc $x=\frac{2}{3}$

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

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

b) Ta có: \(x^2-5x+6=0\)

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

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

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

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

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

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

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

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

f, 3x²+4x-4=0

\(\Leftrightarrow\)3x²+6x-2x-4=0

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

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

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

^{\(\Leftrightarrow\)}\(\left[{}\begin{matrix}x=-2\\x=\frac{2}{3}\end{matrix}\right.\left(tm\right)\)

Vậy pt có tập nghiệm S = \(\left\{-2;\frac{2}{3}\right\}\)

a: =>-2x=-8

hay x=4

b: =>7x=-21

hay x=-3

c: =>0,25x=-1,5

hay x=-6

d: =>5,3x=6,36

hay x=6/5

e: =>-4x=-12

hay x=3

f: =>-10x=-10

hay x=1

g: =>2x+2-3-2x=0

=>-1=0(vô lý)

h: =>3-3x+4x-3=0

=>x=0

a,

\(3-x=x-5\\ \Leftrightarrow3x-x+5=0\Leftrightarrow2x+5=0\)

\(\Rightarrow x=-\dfrac{5}{2}\)

b, \(\Rightarrow x=-\dfrac{21}{7}=-3\)

c, \(\Leftrightarrow x=\left(0-1,5\right):0,25=-6\)

a. <=> 2x=8 hay x=4

b.<=> x= -21/7 = -3

c. <=> x= -1,5/ 0,25=-6

d. <=> x= -6,36/-5,3=1,2

e.<=> 4x=12 hay x= 3

f. <=> 10x = 10 hay x = 1

g. <=> 2x +2 = 3 + 2x

<=> 2=3 ( vô lí )

h.<=> 3 - 3x + 4x -3 =0

<=> x=0

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

\(j,3x^2+7x+2=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

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

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

Vậy...............................

\(m,3x^2+4x-4=0\)

\(\Leftrightarrow3x^2+6x-2x-4=0\)

\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

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

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

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