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phần cuối mk chụp ko đc hết . chỗ cuối là bằng \(\frac{-5}{-3}\)=\(\frac{5}{3}\)
$ (3x-1)(2x+7)-(x+1)(6x-5)=16 $
$ 6x^2+21x-2x-7-(6x^2+x-5)=16 $
$ 25x-7-x+5=16 $
$ 24x-2=16 $
$ 24x=18 $
$ x=\frac{3}{4} $
$ (10x+9)x-(5x-1)(2x+3)=8 $
$ 10x^2+9x-(10x^2+13x-3)=8 $
$ 10x^2+9x-10x^2-13x+3=8 $
$ -4x+3=8 $
$ -4x=5 $
$ x=-\frac{5}{4} $
- Tại câu a có hai bạn ra hai đáp án nên góp thêm lựa ý kiến cho người hỏi đỡ bối rối 
a, Ta có : \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
=> \(\left(x-3\right)\left(x^2-6x+9-x^2-3x-9\right)+9\left(x+1\right)^2=15\)
=> \(9\left(x+1\right)^2-9\left(x^2-3x\right)=15\)
=> \(9\left(x^2+2x+1-x^2+3x\right)=15\)
=> \(9\left(5x+1\right)=15\)
=> \(5x=\frac{15}{9}-1=\frac{2}{3}\)
=> \(x=\frac{2}{15}\)
Vậy ....
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
a,\(3x\left(x-1\right)+x-1=0\)
\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(3x+1\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
c,\(\left(2x-1\right)^2-25=0\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)