1/4 - 5/2 x |3x - 1/5|=2/3 x |3x - 1/5|- 2/3

Tương đương với 1/4+2/3 = 2/3 x l3x - 1/5l + 5/2 x l3x-1/5l

                           11/12    = l3x - 1/5l x (2/3 + 5/2)

                            11/12  = l3x -1/5 l x 19/6

        => l3x - 1/5l = 11/12 : 19/6 = 11/38

  Xét 2 trường hợp:

      + 3x - 1/5 = 11/38 => 3x = 11/38 + 1/5 = 93/190 => x = 93/190 : 3 = 31/190

      + 3x - 1/5 = -11/38 => 3x = -11/38 + 1/5 = -17/190 => x = -17/190 : 3 = -17/570

x²( x + 1 ) + 2x( x + 1 ) = 0 <=> x( x + 1 )( x + 2 ) = 0 <=> x = 0 hoặc x = -1 hoặc x = -2

x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5

Trả lời:

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

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

\(\Leftrightarrow x=0;x=-1;x=-2\)

Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.

2, \(x\left(3x-1\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5\end{cases}}}\)

Vậy x = 1/3; x = - 5 là nghiệm của pt.

\(\Leftrightarrow\left(x+1\right)\left(3x-5\right)=0\)

hay \(x\in\left\{-1;\dfrac{5}{3}\right\}\)

a) \(\left|2x-5\right|=x+1\)

<=> \(\orbr{\begin{cases}2x-5=x+1\left(x\ge\frac{5}{2}\right)\\5-2x=x+1\left(x< \frac{5}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\3x=4\end{cases}}\)

<=> \(x=\frac{4}{3}\left(tm\right)\)

b) \(\left|3x-2\right|-1=2x\) <=> \(\left|3x-2\right|=2x+1\)

<=> \(\orbr{\begin{cases}3x-2=2x+1\left(x\ge\frac{2}{3}\right)\\2-3x=2x+1\left(x< \frac{2}{3}\right)\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-3\left(ktm\right)\\5x=1\end{cases}}\) <=> \(x=\frac{1}{5}\left(tm\right)\)

c) \(\left|x-5\right|+5=x\) <=> \(\left|x-5\right|=x-5\)

<=> \(\orbr{\begin{cases}x-5=x-5\left(x\ge5\right)\\5-x=x-5\left(x< 5\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\\2x=10\end{cases}}\) <=> 0x = 0 (luôn đúng) hoặc x = 5 (ktm)

Vậy x \(\ge\)5

d) \(\left|3x-5\right|=3x-5\) <=> \(\orbr{\begin{cases}3x-5=3x-5\left(x\ge\frac{5}{3}\right)\\5-3x=3x-5\left(x< \frac{5}{3}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\left(luônđúng\right)\\6x=10\end{cases}}\)

<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)Vậy x \(\ge\)5/3

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

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

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

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

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