a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow\left(x^3+2^3\right)-\left(x^3-2x\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x+8=15\) 

\(\Leftrightarrow2x=15-8\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) \(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4=6\)

\(\Leftrightarrow-8x+20=6\)

\(\Leftrightarrow-8x=6-20\)

\(\Leftrightarrow-8x=-14\)

\(\Leftrightarrow x=\dfrac{7}{4}\) 

c) \(x^4-2x^3+x^2-2x=0\)

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

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

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

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

`(x+2)(x^2 -2x+4) -x(x^2-2)=15`

`<=> x^3 +8 - x^3 + 2x-15=0`

`<=> 2x-7=0`

`<=> 2x=7`

`<=>x=7/2`

__

`(x-4)^2 -(x-2)(x+2)=6`

`<=>x^2 - 8x+16- x^2 +4-6=0`

`<=> -8x+14=0`

`<=> -8x=-14`

`<=>x=14/8= 7/4`

__

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

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

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

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

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

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

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

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

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

\(\left|2x+6\right|+\left|x+2\right|+\left|2x+2\right|=4\)

\(\Leftrightarrow2\left|x+3\right|+\left|x+2\right|+2\left|x+1\right|=4\) (1)

+, Với \(x< -3\) thì (1) trờ thành:

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

\(\Rightarrow-2x-6-x-2-2x-2=4\)

\(\Rightarrow-5x-10=4\)

\(\Rightarrow-5x=14\)

\(\Rightarrow x=-\dfrac{14}{5}\left(ktm\right)\)

+, Với \(-3\le x< -2\), thì (1) trở thành:

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

\(\Rightarrow2x+6-x-2-2x-2=4\)

\(\Rightarrow-x+2=4\)

\(\Rightarrow x=-2\left(ktm\right)\)

+, Với \(-2\le x< -1\), thì (1) trở thành:

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

\(\Rightarrow2x+6+x+2-2x-2=4\)

\(\Rightarrow x+6=4\)

\(\Rightarrow x=-2\left(tm\right)\)

+, Với \(x\ge-1\), thì (1) trở thành:

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

\(\Rightarrow2x+6+x+2+2x+2=4\)

\(\Rightarrow5x+10=4\)

\(\Rightarrow5x=-6\)

\(\Rightarrow x=-\dfrac{6}{5}\left(ktm\right)\)

Vậy \(x=-2\)

#Urushi☕

|2x+6|+ |x+2| + |2x+2| =4

=>2x+6+ x+2 + 2x+2 =4 hoặc 2x+6+ x+2 + 2x+2 =-4

=>2x+6+ x+2 + 2x+2 =4         2x+6+ x+2 + 2x+2 =-4

=>(2x+x+2x)+(6+2+2)=4         (2x+x+2x)+(6+2+2)=-4

=> 5x+10=4                              5x+10=-4   

=>5x=4-10                                 5x=(-4)-10     

=>5x=-6                                     5x=-14

=>x=(-6):5                                     x= (-14):5

=>x=-6/5                                         x=-14/5

vậy x∈{-6,5;-14/5}

|2x+6|+|x+2|+|2x+2|=4

=>2|x+3|+|x+2|+2|x+1|=4(1)
TH1: x<-3

(1) sẽ trở thành:

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

=>-2x-6-x-2-2x-2=4

=>-5x-10=4

=>-5x=14

=>x=-14/5(loại)

TH2: -3<=x<-2

Phương trình (1) sẽ là:

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

=>2x+6-x-2-2x-2=4

=>-x+2=4

=>-x=2

=>x=-2(loại)

TH3: -2<=x<-1

Phương trình (1) sẽ là:

2x+6+x+2-2x-2=4

=>x+6=4

=>x=-2(nhận)

Th4: x>=-1

Phương trình (1) sẽ trở thành:

2x+6+x+2+2x+2=4

=>5x+10=4

=>5x=-6

=>x=-6/5(loại)

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

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)

=>24-6x-4x+4=2x+40

=>-10x+28=2x+40

=>-12x=12

=>x=-1

\(6\left(4-x\right)-4\left(x-1\right)=2x+40\)

\(\Rightarrow24-6x-4x+4=2x+40\)

\(\Rightarrow-6x-4x-2x=40-24-4\)

\(\Rightarrow-12=12\)

\(\Rightarrow x=\dfrac{12}{-12}\)

\(\Rightarrow x=-1\)

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6