a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?

1.

a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)

b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)

2.

a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)

b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14

`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)

a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 ) 

<=> 2x - 3 - x + 5 = x + 7 - x - 2

<=> x = 3

b)(7x-5)-(6x+4)=(2x+3)-(2x+1)

<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1

<=> x = 11

c)(9x-3)-(8x+5)=(3x+2)

<=> 9x - 3 - 8x - 5 = 3x + 2

<=> -2x = 10

<=> x = -5

d)(x+7)-(2x+3)=(3x+5)-(2x+4)

<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4

<=> -2x = -3

<=> x = 3/2

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Mài ngu

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

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

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

\(A=x^2+6x+9-4x-1-2x-x^2=9\\ B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ C=\left(3x+5-3x+5\right)^2=100\)

a: \(A=x^2+6x+9-4x-1-2x-x^2=8\)

b: \(B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\)

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

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