Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)

\(< =>2x-7=0< =>x=\dfrac{7}{2}\)

b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)

\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(< =>-25x-11=0\)

\(< =>x=-0,44\)

a.\(x^2-25=8\left(5-x\right)\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)

b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)

\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)

\(\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

a) \(x^2+2x=\left(x-2\right).3x\)

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

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

\(\Leftrightarrow-2x^2+8x=0\)

\(\Leftrightarrow-2x\left(x-4\right)=0\)

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

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

Vậy S = {0;4}

b) \(x^3+x^2-x-1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)

Vậy: S = {-1; 1}

c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt x² + 6x + 5 = t

\(\Leftrightarrow t.\left(t+3\right)=40\)

\(\Leftrightarrow t^2+3t=40\)

\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)

\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)

Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)

=> x² + 6x = 0

<=> x. (x + 6) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy S = {0; -6}

a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)

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

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

\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)

\(\Leftrightarrow x\left(-2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy: S={0;4}

b) Ta có: \(x^3+x^2-x-1=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)=0\)

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

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

Vậy: S={-1;1}

c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)

\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)

\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)

mà \(x^2+6x+13>0\forall x\)

nên \(x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy: S={0;-6}

b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)

a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)

c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)

a)\(\Leftrightarrow\)\(x^2-4x-21=0\)

\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)

\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)

\(\Leftrightarrow\)\((x-7)(x+3)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)

b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)

\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)

\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)

c)\((x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15\)

\(\Leftrightarrow\)\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9-15=0\)

\(\Leftrightarrow\)\(45x-6=0\)

\(\Leftrightarrow\)\(x=\dfrac{2}{15}\)

a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)

=>x(2x+10-x+2)=0

=>x(x+12)=0

=>x=0 hoặc x=-12

b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

=>(2x-5)(3x+12)=0

=>x=5/2 hoặc x=-4

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

=>(x-3)(3x+3)=0

=>x=3 hoặc x=-1

d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

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

=>(x+2)(-5x+3)=0

=>x=-2 hoặc x=3/5

\(a,\left(x-2\right)x=2x\left(x+5\right)\)

\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)

\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)