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1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(d,x\left(x-3\right)-7x+21=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
\(a,2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
\(c,4x^2+12x+9=0\)
\(\Leftrightarrow4x^2+6x+6x+9=0\)
\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\\ =\left(x^2+10x\right)\left(x^2+10x+24\right)+128\\ =\left(x^2+10x+12-12\right)\left(x^2+10x+12+12\right)+128\\ =\left(x^2+10x+12\right)-12^2+128=\left(x^2+10x+12\right)^2-16\\ =\left(x^2+10x+12-16\right)\left(x^2+10x+12+16\right)\\ =\left(x^2+10-4\right)\left(x^2+10x+28\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
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a/ ĐKXĐ: \(x\ne2;3\)
\(\dfrac{x+3}{x-2}+\dfrac{5}{\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x-3\right)+5}{\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x^2-9+5=0\Leftrightarrow x^2=4\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne\pm\dfrac{3}{4}\)
\(\dfrac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}+\dfrac{3x-7}{4x-3}-\dfrac{6x+5}{4x+3}=0\)
\(\Leftrightarrow12x^2+30x-21+\left(3x-7\right)\left(4x+3\right)-\left(6x+5\right)\left(4x-3\right)=0\)
\(\Leftrightarrow9x-27=0\Rightarrow x=3\)
c/ ĐKXĐ: \(x\ne-1;2\)
\(\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}-\dfrac{4}{x+1}+\dfrac{2}{x-2}=0\)
\(\Leftrightarrow x+3-4\left(x-2\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow-x+13=0\)
\(\Rightarrow x=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)
\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm
5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)
6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)
\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)
(12x + 4)/21 - (x - 3)/3 = (3(x - 2))/7
<=> (4(3x + 1))/21 - (x - 3)/3 = (3(x - 2))/7
<=> 4(3x + 1) - 7(x - 3) = 9(x - 2)
<=> 12x + 4 - 7x + 21 = 9x - 18
<=> 5x + 25 = 9x - 18
<=> 5x + 25 - 9x = -18
<=> -4x + 25 = -18
<=> -4x = -18 - 25
<=> -4x = -43
<=> x = 43/4