a: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

=>\(x^2+8x+16-\left(x^2-1\right)=16\)

=>\(x^2+8x+16-x^2+1=16\)

=>8x+17=16

=>8x=-1

=>\(x=-\frac18\)

b: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

=>\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

=>\(5x^2+2x+10-5x^2+245=0\)

=>2x+255=0

=>2x=-255

=>\(x=-\frac{255}{2}\)

c: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

=>\(x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

=>\(x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

=>24x+25=49

=>24x=24

=>x=1

a: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

=>\(x^2+8x+16-\left(x^2-1\right)=16\)

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

=>8x+1=0

=>8x=-1

=>\(x=-\frac18\)

b: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\cdot\left(x-7\right)=0\)

=>\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

=>\(5x^2+2x+10-5x^2+245=0\)

=>2x+255=0

=>2x=-255

=>\(x=-\frac{255}{2}\)

c: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

=>\(x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

=>\(x^3-6x^2+12x-8-x^3+27+6x^2+12x+6\) =49

=>24x+25=49

=>24x=24

=>x=1

\(a,\left(x+7\right)\left(3x-1\right)=x^2-49\)

\(\left(x+7\right)\left(3x-1\right)-\left(x+7\right)\left(x-7\right)=0\)

\(\left(x+7\right)\left(3x-1-x+7\right)=0\)

\(\left(x+7\right)\left(2x+6\right)=0\)

\(\hept{\begin{cases}x+7=0\\2x+6=0\end{cases}}\)

\(\hept{\begin{cases}x=-7\\x=-3\end{cases}}\)

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

\(5x-15-4=2x-2+7\)

\(5x-19=2x+5\)

\(3x=24\)

\(x=8\)

\(a.\left(x+7\right)\left(3x-1\right)=x^2-49\Leftrightarrow\left(x+7\right)\left(3x-1\right)=\left(x-7\right)\left(x+7\right)\Rightarrow3x-1=x-7\Rightarrow3x-x=-7+1\Rightarrow2x=-6\Rightarrow x=-3\)

Ko chép đề

\(2)x-17+x=x-7\)

\(x+x-x=-7+17\)

\(x=-10\)

Vậy ....

âm 10 , nhỉ ?

a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)

\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)

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

\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)

c,  thiếu đề 

d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)

\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)

a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)

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

b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)

\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)

c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)

\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)

d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)

giờ làm vẫn đc đúng ko bạn