\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

Đề mik ko hiểu lắm

\(2x\left(x-4\right)-6x^2\left(4-x\right)=0\)

\(\Leftrightarrow6x^2\left(x-4\right)+2x\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)\left(3x+1\right)=0\)

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

x=\(\dfrac{15}{11}\)

=>-4x+12=6x+x-3

=>3x-3=-4x+12

=>7x=15

hay x=15/7

<=> x² -4+3x²= 4x²+4x+1+2x

<=> 4x^2 - 4= 4x^2 +6x +1

<=> - 4=6x +1

<=> 6x= -5

<=> x= \(-\frac{5}{6}\)

7(x - 3) - x(3 - x)

= (x - 3)(7 + x)

chỉ bt có v mà k bt có đúng k 

1 ) 7 ( x - 3 ) - x ( 3 - x ) 

= 7 ( x - 3 ) + x ( x - 3 )

= ( x - 3 ) ( 7 + x )

2 ) 4x² - 6x + 3 - 2x

= 4x² - 2x - 6x + 3

= 2x ( 2x - 1 ) - 3 ( 2x - 1 )

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

3 ) ( 4 - x ) - 4x + x²

= ( 4 - x ) - x ( 4 - x )

= ( 4 - x ) (  1 - x )

4 ) x² - 2xy + y²

= ( x - y )²

a)(x+3)2-x2+15=2x+6-2x+15=1

                        =21=1

Bạn chép sai đầu bài à

x2 là x^2 à

Ukm

It's very hard

l can't do it 

Sorry!

a) \(x^4-x^3-7x^2+x+6=0\)

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

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

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

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

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

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

b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)

\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)

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

Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)

\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)

\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)

Từ đó tính đc x

d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+5=a\), khi đó pt có dạng:

\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)