Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

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

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) = x³ + 9x² + 27x + 27 - 9x³ -6x² - x + 8x³ +1 -3x² =54

26x +28 = 54

26x = 54-28 = 26

x = 1

b) = x³ - 9x² + 27x -27 - x³ +27 +6x² + 12x + 6 +3x² = -33

39x +6 = -33

39x = -33-6 = -39

x = -1

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

\(8x-48+4x-12-14=-x+4\)

\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)

\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)

\(\Leftrightarrow2x=130\Leftrightarrow x=65\)

\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)

\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)

\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)

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

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

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

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

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

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

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

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

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

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

TYUIUYGBGJCYYHBVJHBNGHJK.;;JHGFFDSAQWERTYUIO

Bài 1:

a, 96 \(⋮x=>x\inư\left(96\right)\)

b, \(2^x.15+2^x.17=4^{30}\)

\(2^x\left(15+17\right)=4^{30}\)

\(2^x.32=4^{30}\)

\(2^x.2^5=2^{60}\)

\(2^x=2^{60}:2^5\)

\(2^x=2^{55}\)

=> x = 55

a) 4( 18 - 5x ) - 12( 3x - 16 ) = 15( 2x - 16 ) - 6( x + 14 )

<=> 72 - 20x - 36x + 192 = 30x - 240 - 6x - 84

<=> -20x - 36x - 30x + 6x = -240 - 84 - 72 - 192

<=> -80x = -588

<=> x = -588/-80 = 147/20

b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 6

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 6

<=> x² + 5x + 6 - x² - 3x + 10 = 6

<=> 2x + 16 = 6

<=> 2x = -10

<=> x = -5

c) -x( x + 3 ) + 2 = ( 4x + 1 )( x - 1 ) + 2x

<=> -x² - 3x + 2 = 4x² - 3x - 1 + 2x

<=> -x² - 3x - 4x² + 3x - 2x = -1 - 2

<=> -5x² - 2x = -3

<=> -5x² - 2x + 3 = 0

<=> -( 5x² + 2x - 3 ) = 0

<=> -( 5x² + 5x - 3x - 3 ) = 0

<=> -[ 5x( x + 1 ) - 3( x + 1 ) ] = 0

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

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

d) ( 2x + 3 )( x - 3 ) - ( x - 3 )( x + 1 ) = ( 2 - x )( 3x + 1 ) + 3 

<=> 2x² - 3x - 9 - ( x² - 2x - 3 ) = -3x² + 5x + 2 + 3

<=> 2x² - 3x - 9 - x² + 2x + 3 = -3x² + 5x + 2 + 3

<=> 2x² - 3x - x² + 2x + 3x² - 5x = 2 + 3 + 9 - 3

<=> 4x² - 6x = 11

<=> 4x² - 6x - 11 = 0

=> Vô nghiệm ( Lớp 8 chưa học nghiệm vô tỉ nên để vậy ) :))

vẫn làm được nha quỳnh !

\(4x^2-6x-11=0\)

\(< =>\left(4x^2-6x+\frac{9}{4}\right)-13\frac{1}{4}=0\)

\(< =>\left(2x-\frac{3}{2}\right)^2=\frac{53}{4}\)

\(< =>\orbr{\begin{cases}2x-\frac{3}{2}=\frac{\sqrt{53}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}2x=\frac{3+\sqrt{53}}{2}\\2x=\frac{3-\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{3+\sqrt{53}}{4}\\x=\frac{3-\sqrt{53}}{4}\end{cases}}\)

:vc dcv ơi, lớp 8 chưa học số vô tỉ ớ nên cho PT đó vô nghiệm là đúng, cái j mà ko làm đc:)) haha ( pịt momx lại ko bị chửi mù chữ:)

a, \(4\left(18-5x\right)-12\left(3x-16\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Leftrightarrow72-20x-36x+192=30x-240-6x-84\)

\(\Leftrightarrow264-56x=24x-324\Leftrightarrow x=\frac{147}{20}\)

c, \(-x\left(x+3\right)+2=\left(4x+1\right)\left(x-1\right)+2x\)

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

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

\(\Leftrightarrow5x^2+2x-3=0\Leftrightarrow\left(x+1\right)\left(5x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

x=-15/6