Trả lời:

x = -3

Mk ms lp 6 à. Nên còn cách nào nx hay ko thì chịu

Băng'ss Băng'ss

\(|x+1|=|2x+3|\)

\(\Rightarrow\left(x+1\right)^2=\left(2x+3\right)^2\)

\(\Rightarrow x^2+2x+1=4x^2+12x+9\)

\(\Rightarrow3x^2+10x+8=0\)

Giải phương trình bậc 2 ....

<=> x+1=2x+3 hoặc x+1=-2x-3

<=> x-2x=3-1  hoặc x+2x=-3-1

<=>-x=2 hoặc 3x=-4

<=> x=-2 hoặc x=-4/3

Không biết nãy bị lỗi ở đâu, mình gửi lại:<

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a  A=4x-x^2+3

      =(x-2)^2-1

     MIN A= -1 khi (x-2)^2=0

      x-2=0

      x=2

B=x-x^2

B=-x^2+x

-B=x^2-x

-B=(x-1/2)^2-1/4

B=-(x-1/2)^2+1/4

MAX B=1/4 khi -(x-1/2)^2=0

x-1/2=0

x=1/2

N=2x-2x^2-5

-N=2x^2-2x+5

-N=2(x^2-x+2)+1

-N=2{(x-1/2)^2+7/4}+1

-N=2(x-1/2)^2+7/2+1

-N=2(x-1/2)^2+9/2

N=-2(x-1/2)^2-9/2

MAX N=-9/2 khi -2(x-1/2)^2=0

x-1/2=0

x=1/2

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

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

Đặt \(x^2+2x=t\)

pt <=> \(t^2-2t=4\)

\(\Leftrightarrow t^2-2t-4=0\)

...

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

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

Đặt \(x^2+2x=a\)

\(\Rightarrow pt\Leftrightarrow a^2-2a=4\Leftrightarrow a^2-2a-4=0\)

\(\cdot\Delta=\left(-2\right)^2-4.\left(-4\right)=20,\sqrt{\Delta}=\sqrt{20}\)

Vậy pt ẩn phụ có 2 nghiệm phân biệt

\(a_1=\frac{2+\sqrt{20}}{2}=\sqrt{5}+1\);\(a_2=\frac{2-\sqrt{20}}{2}=1-\sqrt{5}\)

Thay vào \(x^2+2x=a\),dùng delta giải. 

e: \(\dfrac{12-3x}{2x+6}-3>0\)

\(\Leftrightarrow\dfrac{12-3x-6x-18}{2x+6}>0\)

\(\Leftrightarrow\dfrac{-9x+6}{2x+6}>0\)

\(\Leftrightarrow\dfrac{3x-2}{x+3}< 0\)

=>-3<x<2/3

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10