ĐKXĐ \(3x^2-5x+1\ge0;x^2-2\ge0;x^2-x-1\ge0\)

Ta có : \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3.\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

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

\(\Leftrightarrow\dfrac{3x^2-5x+1-3.\left(x^2-x-1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}=\dfrac{x^2-2-x^2+3x-4}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}+\dfrac{2}{\sqrt{3x^2-5x+1}+\sqrt{3\left(x^2-x-1\right)}}=0\left(∗\right)\end{matrix}\right.\)

Xét phương trình (*) ta có VT > 0 \(\forall x\) mà VP = 0

nên (*) vô nghiệm

Vậy x = 2 là nghiệm phương trình 

a: Đặt \(x^2-4=a\)

Pt sẽ là \(a=3\sqrt{xa}\)

\(\Rightarrow a^2=9xa\)

\(\Leftrightarrow a\left(a-9x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)

hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)

d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)

Pt sẽ là 2a+b=ab+2

=>(b-2)(1-a)=0

=>b=2 và 1-a

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

Điều kiện xác định phương trình \(x\ge\frac{1}{4}\).

Nhân cả hai vế với \(\sqrt{2}\) phương trình tương đương với

\(\sqrt{4x-2\sqrt{4x-1}}-\sqrt{4x+2\sqrt{4x-1}=4}\leftrightarrow\left|\sqrt{4x-1}-1\right|-\left|\sqrt{4x-1}+1\right|=4\)

\(\leftrightarrow\left|\sqrt{4x-1}-1\right|-\sqrt{4x-1}=5\).

Trường hợp 1. NẾU \(x\ge\frac{1}{2}\to\sqrt{4x-1}-1-\sqrt{4x-1}=5\to\) loại

Trường hợp 2. NẾU \(\frac{1}{4}\le x

Giải bằng bất đẳng thức Cô si: (ĐK: \(x^2-x+1\ge0;-2x^2+x+2\ge0;x^2-4x+7\)
Ta có: \(x^2-x+1+1\ge2\sqrt{x^2-x+1}\Leftrightarrow\sqrt{x^2-x+1}\le\dfrac{x^2-x+2}{2}\left(1\right)\\ T,T:\sqrt{-2x^2+x+2}\le\dfrac{-2x^2+x+3}{2}\left(2\right)\\ \left(1\right);\left(2\right)\Rightarrow\sqrt{x^2-x+1}+\sqrt{-2x^2+x+2}\le\dfrac{x^2-x+2-2x^2+x+3}{2}=\dfrac{-x^2+5}{2}\\ \Rightarrow\sqrt{x^2-x+1}+\sqrt{-2x^2+x+2}-\dfrac{x^2-4x+7}{2}\le\dfrac{-x^2+5-x^2+4x-7}{2}\\ =\dfrac{-2x^2+4x-2}{2}\\ =-x^2+2x-1 \\ \Rightarrow-\left(x-1\right)^2\ge0\)
Điều này chỉ thỏa 1 điều kiên khi x-1=0 ⇔x=1(nhận
Vậy x=1 là nghiệm cuả phương trình

Câu 3: đề là \(\sqrt{x+5}-\sqrt{x-2}\) hay \(\sqrt{x+5}-\sqrt{x+2}\)?

Câu 4:

ĐKXĐ: \(x\le9\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{9-x}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a-b=-1\\a^3+b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\a^3+b^2=5\end{matrix}\right.\)

\(\Rightarrow a^3+\left(a+1\right)^2=5\)

\(\Leftrightarrow a^3+a^2+2a-4=0\) \(\Rightarrow a=1\)

\(\Rightarrow\sqrt[3]{x-4}=1\Rightarrow x-4=1\Rightarrow x=5\)

5.

ĐKXĐ: \(x\ge-\frac{17}{16}\)

\(\Leftrightarrow8x^2-15x-23-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left(x+1\right)\left(8x-23\right)-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8x-23=\sqrt{16x+17}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow16x+17-2\sqrt{16x+17}-63=0\)

Đặt \(\sqrt{16x+17}=t\ge0\)

\(\Rightarrow t^2-2t-63=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{16x+17}=9\Leftrightarrow x=\frac{32}{3}\)

mình cần phần 3 4 5 nữa thui ạ

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

ĐKXĐ : \(x\ge5\)

Ta có \(x-3\sqrt{x}+4=2\sqrt{x-5}\)

\(\Leftrightarrow x-3\sqrt{x}=2\left(\sqrt{x-5}-2\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)=2.\dfrac{x-9}{\sqrt{x-5}+2}\)

\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-3\right)=\dfrac{2\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}{\sqrt{x-5}+2}\)

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

Với \(\sqrt{x}-3=0\Leftrightarrow x=9\left(tm\right)\)

Với \(\sqrt{x}=\dfrac{2.\left(\sqrt{x}+3\right)}{\sqrt{x-5}+2}\Leftrightarrow\sqrt{x}.\sqrt{x-5}=6\)

\(\Leftrightarrow x^2-5x-36=0\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=-4\left(\text{loại}\right)\end{matrix}\right.\)

Tập nghiệm \(S=\left\{9\right\}\)