\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{2x-9}=\sqrt[3]{x-5}+\sqrt[3]{4x-3}\)

Đặt \(\sqrt[3]{3x+1}=a;\sqrt[3]{2x-9}=b;\sqrt[3]{x-5}=c;\sqrt[3]{4x-3}=d\) ta được hệ:

\(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=c+d\\\left(a+b\right)^3-3ab\left(a+b\right)=\left(c+d\right)^3-3cd\left(c+d\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+b=c+d=0\\\left[{}\begin{matrix}a+b=c+d\ne0\\ab=cd\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a^3+b^3=0\\a^3b^3=c^3d^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\\left(3x+1\right)\left(2x-9\right)=\left(4x-3\right)\left(x-5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-8=0\\x^2-x-12=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(3x^2+\sqrt{2}x-3+\sqrt{2}=0\)

Ta có \(a-b+c=3-\sqrt{2}-3+\sqrt{2}=0\)

Vậy phương trình có 2 nghiệm phân biệt

\(x_1=-1\)

\(x_2=-\dfrac{-3+\sqrt{2}}{3}=\dfrac{3-\sqrt{2}}{3}\)

a/ \(\Delta=\left(3\sqrt{3}\right)^2-4.4\left(-6\right)=123\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3\sqrt{3}+\sqrt{123}}{8}\\x_2=\frac{3\sqrt{3}-\sqrt{123}}{8}\end{matrix}\right.\)

b/ \(\Delta=9-4\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)=25\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3+\sqrt{25}}{2\left(1-\sqrt{5}\right)}=-1-\sqrt{5}\\x_2=\frac{3-\sqrt{25}}{2\left(1-\sqrt{5}\right)}=\frac{1+\sqrt{5}}{4}\end{matrix}\right.\)

\(a)4x^2-3\sqrt{3}x-6=0\)

Có: \(a=4;b=-3\sqrt{3};c=-6\)

\(\Delta=b^2-4ac\\ =\left(-3\sqrt{3}\right)^2-4.4.\left(-6\right)\\ =123>0\)

Phương trình có 2 nghiệm phân biệt:

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-3\sqrt{3}\right)+\sqrt{123}}{2.4}=\frac{3\sqrt{3}+\sqrt{123}}{8}\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-3\sqrt{3}\right)-\sqrt{123}}{2.4}=\frac{3-\sqrt{123}}{8}\)

\(b)\left(1-\sqrt{5}\right)x^2-3x+\sqrt{5}+1=0\)

Có: \(a=1-\sqrt{5};b=-3;c=\sqrt{5}+1\)

\(\Delta=b^2-4ac\\ =\left(-3\right)^2-4.\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\\ =25>0\)

Phương trình có 2 nghiệm phân biệt:

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)+\sqrt{25}}{2\left(1-\sqrt{5}\right)}=-1-\sqrt{5}\\ x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)-\sqrt{25}}{2\left(1-\sqrt{5}\right)}=\frac{1+\sqrt{5}}{4}\)

Nguyễn Việt Lâm giúp mk nhá, thanks bn nhìu :>>>

a, ĐKXĐ: ...

\(\sqrt{3x^2-2x+6}+3-2x=0\)

\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

\(\sqrt{3x^2}-\left(1-\sqrt{3}\right)x-1=0\)

\(\Leftrightarrow\sqrt{3}x-x-\sqrt{3}x-1=0\)

\(\Leftrightarrow-x-1=0\)

\(\Leftrightarrow-x=1\)

\(\Leftrightarrow x=-1\)

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé