\(\dfrac{3x}{x^2-4x+7}+\dfrac{2x}{x^2-6x+7}=2\) (x \(\ne\) 3 + \(\sqrt{2}\); x \(\ne\) 3 - \(\sqrt{2}\))

Đặt x² - 5x + 7 = t (t \(\ne\) \(\pm\) x)

Khi đó:

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

\(\Leftrightarrow\) \(\dfrac{3x\left(t-x\right)+2x\left(t+x\right)}{t^2-x^2}=2\)

\(\Leftrightarrow\) 3xt - 3x² + 2xt + 2x² = 2(t² - x²)

\(\Leftrightarrow\) 5xt - x² = 2t² - 2x²

\(\Leftrightarrow\) 2t² - x² - 5xt = 0

\(\Leftrightarrow\) 2(t² - \(\dfrac{5}{2}\)xt + \(\dfrac{25}{16}\)x² - \(\dfrac{33}{16}\)x²) = 0

\(\Leftrightarrow\) (t - \(\dfrac{5}{4}\))² - \(\dfrac{33}{16}\)x² = 0

\(\Leftrightarrow\) (t - \(\dfrac{5}{4}\) - \(\dfrac{\sqrt{33}}{4}\))(t - \(\dfrac{5}{4}\) + \(\dfrac{\sqrt{33}}{4}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}t=\dfrac{5+\sqrt{33}}{4}\\t=\dfrac{5-\sqrt{33}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x^2-5x+7=\dfrac{5+\sqrt{33}}{4}\\x^2-5x+7=\dfrac{5-\sqrt{33}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}+\dfrac{3}{4}=\dfrac{5+\sqrt{33}}{4}\\x^2-2.\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\dfrac{5-\sqrt{33}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left(x-\dfrac{5}{2}\right)^2=\dfrac{2+\sqrt{33}}{4}\\\left(x-\dfrac{5}{2}\right)^2=\dfrac{2-\sqrt{33}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{2+\sqrt{33}}}{2}\\x-\dfrac{5}{2}=\dfrac{\sqrt{2-\sqrt{33}}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{\sqrt{2+\sqrt{33}}+5}{2}\left(TM\right)\\x=\dfrac{\sqrt{2-\sqrt{33}}+5}{2}\left(KTM\right)\end{matrix}\right.\)

Vậy S = {\(\dfrac{\sqrt{2+\sqrt{33}}+5}{2}\)}

Chúc bn học tốt! (Ko bt đúng ko nhưng nhìn số ko đẹp lắm :v)

ĐKXĐ: ....

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\dfrac{3}{x+\dfrac{7}{x}-4}+\dfrac{2}{x+\dfrac{7}{x}-6}=2\)

Đặt \(x+\dfrac{7}{x}-6=t\)

\(\Rightarrow\dfrac{3}{t+2}+\dfrac{2}{t}=2\Leftrightarrow3t+2\left(t+2\right)=2t\left(t+2\right)\)

\(\Leftrightarrow2t^2-t-4=0\)

\(\Leftrightarrow...\)

1/ \(\dfrac{5}{3}\le x\le\dfrac{7}{3}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x-5}=a>0\\\sqrt{7-3x}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\17-6x=2b^2+3\\6x-7=2a^2+3\end{matrix}\right.\)

Mặt khác theo BĐT Bunhiacốpxki:

\(a+b=\sqrt{3x-5}+\sqrt{7-3x}\le\sqrt{\left(1+1\right)\left(3x-5+7-3x\right)}=2\)

\(\Rightarrow0< a+b\le2\)

Ta được hệ pt:

\(\left\{{}\begin{matrix}a^2+b^2=2\\\left(2b^2+3\right).a+\left(2a^2+3\right)b=2+8ab\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2ab^2+3a+2a^2b+3b-8ab-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-2\\2ab\left(a+b\right)+3\left(a+b\right)-8ab-2=0\end{matrix}\right.\)

\(\Rightarrow\left(\left(a+b\right)^2-2\right)\left(a+b\right)+3\left(a+b\right)-4\left(a+b\right)^2+6=0\)

\(\Leftrightarrow\left(a+b\right)^3-4\left(a+b\right)^2+\left(a+b\right)+6=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b=-1< 0\left(l\right)\\a+b=2\\a+b=3>2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow a+b=2\) , dấu "=" xảy ra khi và chỉ khi:

\(3x-5=7-3x\Rightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

2/ ĐKXĐ: \(x\ne\pm2\)

\(\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-\left(\dfrac{15}{x^2-4}+5\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-5.\left(\dfrac{x^2-1}{x^2-4}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{x^2-1}{x^2-4}\right)-4\left[\left(\dfrac{x^2-1}{x^2-4}\right)-\left(\dfrac{x+1}{x-2}\right)^2\right]=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)-4\left(\dfrac{x+1}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}-\dfrac{4\left(x+1\right)}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{4\left(x+1\right)}{x-2}\\\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=4\left(x^2+3x+2\right)\\x^2-3x+2=x^2+3x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2+15x+6=0\\6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5+\sqrt{17}}{2}\\x=\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}+\left(x+1\right)^2=6\)

Mà \(\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

\(\sqrt{5\left(x+1\right)^2+16}\ge\sqrt{16}=4\)

\(\left(x+1\right)^2\ge0\)

\(\Rightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}+\left(x+1\right)^2\ge6\) với mọi x thuộc R.

Dấu "=" xảy ra khi và chỉ khi \(\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Kết luận: \(x=-1.\)

x=-3 đúng thì **** giùm nha bạn

\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)

\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)

Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

đex ~ vừa thấy trên face lướt qua luôn

Akai Haruma

Nguyễn Việt Lâm

<=> x³ + 3x² + 3x + 1 = 0

<=> (x+1)³ = 0 

<=> x+ 1 = 0 

<=> x = -1

PT có nghiệm là x = -1

đặt t = x + 1. Phương trình có dạng:

(6t + 1)².t .(3t +1) = 6

<=> (36t² + 12t + 1).(3t² + t)  = 6

<=> [12.(3t² + t) + 1](3t² + 1) = 6

<=> 12.(3t² +1)² + (3t² +1) - 6 = 0

<=> 12.(3t² +1)² + 9(3t² +1) - 8.(3t² +t)  - 6 = 0

<=> 3(3t² + t). [4(3t² +t) +3] - 2. [4(3t² +t) +3] = 0

<=> [4(3t² +t) +3]. [3(3t² +t) - 2] = 0

<=> 4(3t² +t) +3 = 0 hoặc 3(3t² +t) - 2 = 0

+)  4(3t² +t) +3 = 0 <=> 12t² + 4t + 3 = 0  Vô nghiệm vì 12t² + 4t + 3 = 8t² + (2t +1)² + 2 > 0 với mọi t

+) 3(3t² +t) - 2 = 0 <=> 9t² + 3t - 2 = 0 <=> 9t² + 6t - 3t - 2 = 0 <=> (3t + 2)(3t -1) = 0

=> t = -2/3 hoặc t = 1/3

=> x + 1 = -2/3 hoặc x + 1 = 1/3

=> x = -5/3 hoặc x = -2/3

Cách 1:

GPT :\(5\sqrt{x-1}-\sqrt{x+7}=3x-4\) - Hoc24

Cách 2:

Đặt \(\left\{{}\begin{matrix}\sqrt{25x-25}=a\\\sqrt{x+7}=b\end{matrix}\right.\)  \(\Rightarrow3x-4=\dfrac{a^2-b^2}{8}\)

Pt trở thành:

\(a-b=\dfrac{a^2-b^2}{8}\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-8\right)=0\)

\(\Leftrightarrow...\)

mình giải bằng casio ra x = 0,767591877

sao bạn lại có chữ hiệp sĩ ở bên cạnh tên vậy?

sao vậy bạn

k mk nha

Em thử ạ!

Đặt \(\sqrt[3]{3x^2-x+2011}=a;\sqrt[3]{3x^3-7x+2002}=b;\sqrt[3]{6x-2003}=c\)

Thì được: \(a^3-b^3-c^3=2002\) (1)

Mặt khác theo đề bài \(\left(a-b-c\right)^3=2002\) (2)

Từ (1) và (2) ta được: \(a^3-b^3-c^3-\left(a-b-c\right)^3=0\)

\(\Leftrightarrow3\left(b-a\right)\left(c-a\right)\left(c+b\right)=0\)

\(\Leftrightarrow a=b\text{ hoặc: }c=a\text{ hoặc }c+b=0\)

+) Với a=  b thì \(a^3=b^3\Leftrightarrow3x^2-x+2001=3x^2-7x+2002\)

\(\Leftrightarrow6x-1=0\Leftrightarrow x=\frac{1}{6}\)

... Anh làm tiếp thử ạ.