a) \(\Delta=169-56=113>0\)

\(\hept{\begin{cases}x_1=\frac{13+\sqrt{113}}{14}\\x_2=\frac{13-\sqrt{113}}{14}\end{cases}}\)

b)  \(\Delta=25-4.3.60< 0\)

vô nghiệm 

Pt tương đương:

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt[3]{4x-3}-\sqrt[3]{3x+1}=\sqrt[3]{5-x}+\sqrt[3]{2x-9}=0\left(1\right)\\3\sqrt[3]{-12x^2+5x+3}=3\sqrt[3]{-2x^2+19x-45}\left(2\right)\end{cases}}\)

(1)<=>4x-3=3x+1 và x-5=2x-9<=>x=4

(2)<=>-12x²+5x+3=-2x²+19x-45<=>-5x²-7x+24=0<=>x=8/5 và x=-3

 bạn thử các giá trị x=4,x=8/5 và x=-3 vào pt và kết luận

mik ko hieu vi sao ban suy ra duoc (1) va (2)

bn co the viet ro ra duoc ko ?

theo mik thay thi 2 pt do dau co tuong duong

Mình chuyển vế rồi lập phương, do  4x-3-(3x+1)=2x-9+(5-x) nên mình giản bỏ luôn, hơi tắc xíu

TXĐ: \(x\ge0\)

Phương trình đã cho tương đương:

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

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

\(\Leftrightarrow\left(x-1\right)\left(1+\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}\right)=0\)

\(\Leftrightarrow x-1=0\) (do \(1+\dfrac{1}{\sqrt{2x+1}+\sqrt{3x}}>0\) \(\forall x\ge0\))

\(\Leftrightarrow x=1\)

\(\sqrt{2x+1}-\sqrt{3x}=x-1\)

Điều kiện : x\(\ge0\)

\(\Leftrightarrow\sqrt{2x+1}=x-1+\sqrt{3x}\)

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

\(\Leftrightarrow2x+1=\left(x-1\right)^2+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1=x^2-2x+1+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1-x^2-x-x-2\left(x-1\right)\sqrt{3x}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-2\sqrt[]{3x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-\sqrt{x}\left(\sqrt{x}+2\sqrt{3}\right)=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x\in\varnothing\end{matrix}\right.\) Vậy pt tập nghiệm S={1;0}

\(\sqrt{5-x^6}=\sqrt[3]{3x^4-2}+1\) 

Xét \(\left|x\right|=1\Leftrightarrow\sqrt{5-1}=\sqrt[3]{3-2}+1\)(đúng) 

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\) 

Xét \(\left|x\right|>1\Rightarrow\sqrt{5-x^6}< \sqrt[3]{3x^4-2}+1\)(loại) 

Xét \(\left|x\right|< 1\Rightarrow\sqrt{5-x^6}>\sqrt[3]{3x^4-2}+1\)(loại) 

Vậy Pt có nghiệm (1;-1)

\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)(đk: \(x\ge-\frac{1}{3}\))(1)

đặt \(\sqrt{3x+1}=a\ge0\)

khi đó:

(1) \(\Leftrightarrow\frac{a^2-1}{a+9}=a-1\)

\(\Leftrightarrow a^2-1=a^2+8a-9\)

\(\Leftrightarrow8a=8\Leftrightarrow a=1\left(tm\right)\)

\(\Leftrightarrow\sqrt{3x+1}=1\Leftrightarrow x=0\left(tm\right)\)

vậy x=0 là nghiệm của phương trình

* mk k chắc lắm đâu có j sai bạn sửa cho mk nhé

Đặt \(\sqrt{x+1}=a\)                                            \(ĐKXĐ:x\ge0\)

        \(\sqrt{3x}=b\)                                               

Ta có: \(a-b=b^2-a^2\)

\(\Leftrightarrow a-b+a^2-b^2=0\)

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

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

Mà \(a+b+1>0\forall x\)

\(\Rightarrow a-b=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x+1}=\sqrt{3x}\)

\(\Leftrightarrow x+1=3x\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)

\(ĐKXĐ:x\ge0\)

Ta có PT \(\Leftrightarrow\sqrt{x+1}-\sqrt{3x}-\left(2x-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-\frac{\sqrt{6}}{2}\right)-\left(\sqrt{3x}-\frac{\sqrt{6}}{2}\right)-\left(2x-1\right)=0\)

\(\Leftrightarrow\frac{x+1-\frac{6}{4}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3x-\frac{6}{4}}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-\left(2x-1\right)=0\)

\(\Leftrightarrow\frac{x-\frac{1}{2}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3\left(x-\frac{1}{2}\right)}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\left(x-\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{1}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\right)=0\)

\(\Rightarrow x=\frac{1}{2}\)(TMĐKXĐ)

DDK : \(x\ge1\)

\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

\(\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)

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

\(\Leftrightarrow x-1-3x+2-5x+2=2\sqrt{15x^2-3x-10x+2}\)

\(\Leftrightarrow3-7x=2\sqrt{15x^2-13x+2}\)

\(\Rightarrow9-42x+49x^2=4\left(15x^2-13x+2\right)\)

\(\Leftrightarrow9-42x+49x^2=60x^2-52x+8\)

\(\Leftrightarrow11x^2-10x-1=0\)

\(\Leftrightarrow11x^2-11x+x-1=0\)

\(\Leftrightarrow\left(11x+1\right)\left(x-1\right)=0\)

Giải nốt nha .