\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)

\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)

\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)

b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)

e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)

g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a) x. (2x - 7) - 4x + 14 = 0

⇔ 2x\(^2\) - 7x - 4x + 14 =0

⇔ 2x( x - 2 ) - 7 ( x - 2 ) = 0

⇔ ( 2x - 7 ) ( x - 2 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)

b) x². (x - 1) - 4x + 4 = 0

⇔ x2. (x - 1) - 4( x - 1 ) = 0

⇔(x\(^2\) - 4 ) ( x - 1 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=1\end{matrix}\right.\)

d) 2x3 + 3x2 + 2x + 3 = 0

⇔ 2x( x\(^2\) + 1 ) +3( x\(^2\) + 1 ) = 0

⇔ ( 2x + 3 ) ( x\(^2\) + 1 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\pm1\end{matrix}\right.\)

e) 4x² - 25 - (2x - 5). (2x + 7) = 0

⇔ ( 2x - 5 ) ( 2x + 5 ) - ( 2x - 5 ) (2x + 7 ) = 0

⇔ ( 2x - 5 ) ( 2x + 5 - 2x - 7 ) = 0

⇔-2(2x - 5 ) =0

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

⇔ x= \(\frac{5}{2}\)

g) x³ + 27 + (x + 3). (x - 9) = 0

⇔ ( x+ 3 ) ( x\(^2\) - 3x + 9) + ( x+ 3 ) ( x - 9 ) = 0

⇔ ( x + 3 ) ( x\(^2\) - 3x + 9 + x - 9 ) = 0

⇔ ( x + 3 ) ( x\(^2\) - 2x ) = 0

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

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

^^

a)  x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(x-2)(2x-7)=0

=>x-2=0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

b, x(x-1)+2x-2=0

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

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

=>x+2=0 hoặc x-1=0

=>x=-2 hoặc x=1

c, 2x^3+3x^2+2x+3=0

=>x²(2x+3)+2x+3=0

=>(x²+1)(2x+3)=0

=>x²+1=0 hoặc 2x+3=0

Vì x²+1>0 với mọi x ->vô nghiệm

=>2x+3=0 =>x=-3/2

d, x^3+6x^2+11x+6=0

=>x³+3x³+2x+3x²+3x³+6=0

=>x(x²+3x+2)+3(x²+3x+2)=0

=>(x²+3x+2)(x+3)=0

=>[x²+x+2x+2](x+3)=0

=>[x(x+1)+2(x+1)](x+3)=0

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

=>x+1=0 hoặc x+2=0 hoặc x+3=0

giúp mình với

a)  x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(x-2)(2x-7)=0

=>x-2=0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

b, x(x-1)+2x-2=0

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

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

=>x+2=0 hoặc x-1=0

=>x=-2 hoặc x=1

c, 2x^3+3x^2+2x+3=0

=>x²(2x+3)+2x+3=0

=>(x²+1)(2x+3)=0

=>x²+1=0 hoặc 2x+3=0

Vì x²+1>0 với mọi x ->vô nghiệm

=>2x+3=0 =>x=-3/2

d, x^3+6x^2+11x+6=0

=>x³+3x³+2x+3x²+3x³+6=0

=>x(x²+3x+2)+3(x²+3x+2)=0

=>(x²+3x+2)(x+3)=0

=>[x²+x+2x+2](x+3)=0

=>[x(x+1)+2(x+1)](x+3)=0

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

=>x+1=0 hoặc x+2=0 hoặc x+3=0

=>x=-1 hoặc x=-2 hoặc x=-3

tau cung bui ma chu mi giup tao roi cam on nhe

-_- bài này hôm qua lm rùi

a, \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)

\(\Rightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)

\(\Rightarrow6x^2-30x-6x^2+4x-21x+14=0\)

\(\Rightarrow-47x=-14\Rightarrow x=\dfrac{14}{47}\)

b, \(\left(8x+3\right)x-\left(4x+1\right)\left(2x-7\right)=5\)

\(\Rightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)

\(\Rightarrow8x^2+3x-8x^2+28x-2x+7=5\)

\(\Rightarrow29x=5-7\Rightarrow29x=-2\)

\(\Rightarrow x=\dfrac{-2}{29}\)

c, \(\left(2x-3\right)\left(2x+3\right)-x.\left(4x+1\right)=1\)

\(\Rightarrow\left(2x\right)^2-9-4x^2-x=1\)

\(\Rightarrow4x^2-4x^2-x=1+9\)

\(\Rightarrow-x=10\Rightarrow x=-10\)

Chúc bạn học tốt!!!

a ) \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)

\(\Leftrightarrow6x^2-30x-6x^2+4x-21x+14=0\)

\(\Leftrightarrow-47x+14=0\)

\(\Leftrightarrow x=\dfrac{14}{47}.\)

Vậy ..........................................

b ) \(\left(8x+3\right).x-\left(4x+1\right)\left(2x-7\right)=5\)

\(\Leftrightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)

\(\Leftrightarrow8x^2+3x-8x^2+28x-2x+7=5\)

\(\Leftrightarrow29x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{29}.\)

Vậy.............

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

\(\Leftrightarrow4x^2-9-4x^2-x=1\)

\(\Leftrightarrow-x=10\)

\(\Leftrightarrow x=10\)

Vậy..........

mk ko bt 123

Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10

x-3)^2-2(2x-7)(x-3)=0

Bài 2: 

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

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