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: \(x\ne\pm2\)

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

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

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

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

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

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

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

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.

Akai Haruma

Nguyễn Việt Lâm

\(\frac{-1}{3}\le x\le6\\ \sqrt[]{3x+1}-4-\left(\sqrt[]{6-x}-1\right)+3x^2-14x-5=0\\ \Leftrightarrow\frac{3x-15}{\sqrt[]{3x+1}+4}+\frac{x-5}{\sqrt[]{6-x+1}}+\left(x-5\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(x-5\right)\left(\frac{3}{\sqrt[]{3x+1}}+\frac{1}{\sqrt[]{6-x}+1}+3x-1\right)=0\)

do\(x\ge\frac{-1}{3}\Rightarrow3x+1\ge0\\ \frac{3}{\sqrt[]{3x+1}}+\frac{1}{\sqrt[]{6-x}+1}+3x-1>0\\ \Rightarrow x=5\)

Trả lời :

Con a giai pt vế trái rồi nhân căn bình phương cả 2 vế

Con b cũng giải pt vế phải chuyển vế rồi bình phương cả 2 vế

Chắc vậy

k bt 

\(pt\Leftrightarrow\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)

\(\Leftrightarrow\frac{3x+1-16}{\sqrt{3x+1}+4}+\frac{1-\left(6-x\right)}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}+3x+1\right]=0\)

\(\Leftrightarrow x=5.\)