1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

`a)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

`b)|2x+1|=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

`c)|2x+1|=7`

`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\) 

`d)|2x+5|=|3x-7|`

`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\) 

`e)|2x+7|=1`

`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\) 

`g)|x-2|+|2x-3|=2`

Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`

`pt<=>x-2+2x-3=2`

`<=>3x-5=2`

`<=>3x=7`

`<=>x=7/3(tm)`

Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`

`pt<=>2-x+3-2x=2`

`<=>5-3x=2`

`<=>3x=3`

`<=>x=1(tm)`

Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`

`pt<=>2-x+2x-3=2`

`<=>x-1=2`

`<=>x=3(l)`

`h)|x+2|+|1-x|=3x+2`

Vì `VT>=0=>3x+2>=0=>x>=-2/3`

`=>|x+2|=x+2`

`pt<=>x+2+|1-x|=3x+2`

`<=>|1-x|=2x(x>=0)`

`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\) 

a.

$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=5\\ 2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

b.

$|2x+1|=0$

$\Leftrightarrow 2x+1=0$

$\Leftrightarrow x=-\frac{1}{2}$
c.

$|2x+1|=7$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)

d.

$|2x+5|=|3x-7|$

\(\Leftrightarrow \left[\begin{matrix} 2x+5=3x-7\\ 2x+5=7-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=12\\ x=0,4\end{matrix}\right.\)

e.

$|2x+7|=x-1\Rightarrow x-1\geq 0\Leftrightarrow x\geq 1$
Với $x\geq 1$ thì $|2x+7|=2x+7$

Khi đó pt trở thành:
$2x+7=x-1$

$\Leftrightarrow x=-8< 1$ (vô lý)

Vậy pt vô nghiệm.

g.

$|x-2|+|2x-3|=2$

Nếu $x\geq 2$ thì pt trở thành:

$x-2+2x-3=2$

$\Leftrightarrow 3x-5=2$

$\Leftrightarrow x=\frac{7}{3}$ (thỏa mãn)

Nếu $\frac{3}{2}\leq x< 2$ thì pt trở thành:

$2-x+2x-3=2$

$\Leftrightarrow x=3$ (không thỏa mãn)

Nếu $x< \frac{3}{2}$ thì pt trở thành:

$2-x+3-2x=2$

$\Leftrightarrow 5-3x=2$

$\Leftrightarrow x=1$ (thỏa mãn)

Vậy..........

h.

Từ đề suy ra $x\geq \frac{-2}{3}$

$\Rightarrow |x+2|=x+2$

Nếu  $x\geq 1$ thì $|1-x|=x-1$. PT trở thành:

$x+2+x-1=3x+2$

$\Leftrightarrow 2x+1=3x+2$

$\Leftrightarrow x=-1$ (vô lý)

Nếu $\frac{-2}{3}\leq x< 1$ thì $|1-x|=1-x$. PT trở thành:
$x+2+1-x=3x+2$

$\Leftrightarrow 3=3x+2$

$\Leftrightarrow x=\frac{1}{3}$ (thỏa mãn)

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

P(x)=2x^4+2x^3-5x+3

Q(x)=4x^4-2x^3+2x^2+5x-2

P(x)+Q(x)

=2x^4+2x^3-5x+3+4x^4-2x^3+2x^2+5x-2

=6x^4+2x^2+1

Bài 1:

a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)

\(=12x^2-4x-6x-2-x-3\)

\(=12x^2-11x-5\)

b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)

\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)

\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)

\(=2x^4y+1x^3y^2-2x^2y^3\)

c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)

\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)

\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)

\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)

\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)

\(=-48x^3-8x^2-24x^4+15x-9\)

Bài 1:

a) \(12x^2-11x-5\)

b,c,d tương tự.

b, ( 5/2 - x ) ^2

=25/4-4/5x+x^2

c,( xy/2 - x/3 ) ( xy/2 + x/3)

=(xy/2)^2-(x/3)^2

c: \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

e: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(\dfrac{5}{2}-x\right)^2=\dfrac{25}{4}-5x+x^2\)

c) \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

d) \(\left(3x+\dfrac{2}{3}yz\right)^2=9x^2+4xyz+\dfrac{4}{9}y^2z^2\)

e) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

f) \(\left(2x-y+2\right)^2=\left(2x-y\right)^2+4\left(2x-y\right)+2^2=4x^2+y^2+4-4xy+8x-4y\)

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)

b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y-1\right)\)

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

\(=\left(y+5+3z\right)\left(y+5-3z\right)\)

d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)

\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)

\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)

1:

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

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)

\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)

\(=2y\left(x+y\right)\left(-x-7y\right)\)

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)