K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 5 2018

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

Cauchy-schwarz:

\(\left(1+1+1\right)\left(x^4+x^2+1\right)\ge\left(x^2+x+1\right)^2\)

"="<=>\(x=1\)

b)\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(x^2+x-1=t\)

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow t=\pm5\)

t=5\(\Leftrightarrow x^2+x-1=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

t=-5<=> pt vô nghiệm

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

3 tháng 2 2022

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x=2x^3-16\)

<=>\(8x=-16\)

<=>\(x=-2\)

i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)

<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(6x^2-2x-10=0\)

<=>\(3x^2-x-5=0\)

<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

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

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

3 tháng 2 2022

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)

<=>\(8x=-16\)

<=>x=-2

i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

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

<=>10x=2

<=>x=\(\dfrac{1}{5}\)

5 tháng 3 2020

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

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

\(\Leftrightarrow2x\left(x-1\right)=\left(1-x\right)\left(x+3\right)\)

\(\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)\left(x+3\right)=0\)

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

\(\Rightarrow x=\pm1\)

5 tháng 3 2020

Giúp tớ mấy câu còn lại đi các cậu, tớ cần gấp lắm ạ ;;-;;

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

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

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}